Quadratics

A guide for grade 10 quadratics, including 3 forms and more!

Table of Contents

  • What is quadratics?
  • Parabolas & parts
  • Quadratic vs Linear
  • Vertex form parts
  • Graphing from vertex form (+ step pattern)
  • Finding the vertex given the equation
  • Factored Form + roots
  • Finding the vertex with the roots
  • Graphing factored form
  • Standard Form
  • Standard form graphing
  • Multiplying binomials + special products
  • Algebra tiles
  • Factoring (common, grouping, simple, complex, perfect square trinomial, difference of squares)
  • Completing the square
  • Solving quadratic equations (quadratic formula, factoring, isolating x)
  • Word problems (motion, revenue, area, consecutive numbers, right triangle, optimization)
  • Reflection + an assessment

What are quadratics?

Remember back to grade 9, when we learned about linear equations and how they make straight lines? Quadratic equations make curved lines. The name quadratic is used because quad mean squared, and in a quadratic equation, the variable is squared. There are 3 forms of quadratic equations that we learn in grade 10: vertex form, factored form, and standard form.

Parabola!

A parabola is the graph of a quadratic equation: the curved line mentioned earlier. These lines have a U-shape, and have many parts to it.

PARTS OF A PARABOLA

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The Vertex: The Main Character

A parabola can have a bunch of parts. For one, the vertex. The vertex is the maximum or minimum point of the graph (depending on the graph and equation). This is also where the graph changes directions. Furthermore, the vertex is where the optimal value and the axis of symmetry cross.

Optimal or Minimum/Maximum value

The minimum and maximum value is very similar to the vertex. It is the lowest or the highest point of the parabola, accordingly. This is similar to the vertex because this value is the y coordinate of the vertex. This is also called the optimal value. In the picture to the right, point is a minimum value, because it is the lowest point of the vertex.

Axis of Symmetry

A parabola is symmetrical, meaning if it was split perfectly in half, both sides would be equal. The axis of symmetry would be the line that divides it perfectly in half. This is x coordinate of the vertex.

Y-Intecept

This one's easy. We leaned this back in grade 9. The y-intercept is where the parabola crosses the y-axis.

X-Intercept, Zeros, and Roots

If you know what the y-intercept is, than the x-intercept should be easy for you. The x-intercept is where the parabola touches the x-axis. But be careful. There can be one, two, or even no x-intercepts. X-intercepts are also called zeros and roots, so don't worry if you see these words used interchangeably.

Some Other Interesting Facts About A Parabola

  • A parabola can open up or down. This will be shown in a section below.
  • The vertex is the point where the axis of symmetry and the parabola touch. This is the optimal value (minimum or maximum).

QUADRATIC EQUATION VS LINEAR EQUATION

What's the difference?

In grade 9, we learned about linear relations, and now we're learning about quadratic equations. Many of you might be wondering what the difference is, so here we go!


  • A linear equation creates a straight line, whereas a quadratic equation creates a curved line.
  • In a quadratic equation, the variable is squared.
  • In a linear equation, the first differences are constant. In a quadratic equation, the second differences are constant.


"Wait, wait, wait! Did you just say second differences?"

Yes, I did. In grade 9, we worked with first differences, the differences between two consecutive terms. The second differences are the differences between the first differences. Confusing? Here's a picture to help:

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Check to see if the following relations are quadratic, linear or neither. Remember, linear has constant first difference and quadratic has constant second differences:
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VERTEX FORM

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What Is Vertex Form

Vertex form is the first of three forms of quadratic equations that we learn. This is called vertex form because it is very easy to find the vertex. The x and y coordinates of the vertex are literally in the equation.


An example of vertex form is:

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Here are the parts of the vertex form:

"a" Value

The "a" in the vertex form equation changes the vertical stretch or compression of the parabola. The higher the a value, the narrower the graph. The lower the a value, the wider the graph.


The a value can change a few things in the graph:


  • If a is greater than 0 (meaning positive), then the graph opens upwards. When a is positive and the parabola opens upwards, the vertex is a minimum value.
  • If a is less than 0 (meaning negative), then the graph opens downwards. When a is negative and the parabola opens downwards, the vertex is a maximum value
  • When a is negative, it is called a vertical reflection, or reflection in the x-axis. This is because the graph seems to have reflected and is now facing downwards.


  • If a is greater than 1, then the graph is stretched. This makes the graph narrower.
  • If a is greater than 0, but less than 1 (meaning it's a decimal/fraction), then the graph is compressed. This makes the graph wider.

"h" Value

The "h" in vertex form is responsible for the horizontal translation of the graph. This means that h moves the graph left or right. There are a few things to remember:


  • The h value moves the vertex left or right, and this is called a horizontal translation.
  • The translation always starts at (0,0).


The vertex is moved in the opposite way of the sign.
  • This means that is h is negative, then the vertex will move h units right (which is the positive part of the x-axis).
  • If h is positive, then the vertex will move h units left (which is the negative part of the x-axis).


  • The h is also the x-coordinate of the vertex, but with the opposite sign.

"k" Value

The "k" in the vertex form is used to translate the graph up or down, meaning that it moves the graph up or down. There are a few things to remember about the k value as well, but it's easier than h.


  • The k value moves the vertex up or down, which is called a vertical translation.
  • Like the h value, the vertical translation should start from (0,0). You can put the horizontal and vertical translation together to find out where you must move from (0,0).

Unlike h, the k value moves in the same way as the sign, meaning:
  • If k is positive, then the graph will move k units up.
  • If k is negative, then the graph will move k units down.


  • The k is the y-coordinate of the vertex.

Put them together!

So know we know that a changes the stretch/compression and/or reflection, h changes the horizontal translation, and k changes the vertical translation, we are almost at the point where we can graph.


First, let's try out an example, and see what information we have. We'll use the equation:

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Lets look at the a value:

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The a value is -2.


  • We remember that a changes stretch/compression and reflection.
  • Because a is negative, the parabola is vertically reflected. This means that the parabola opens down.
  • Also, we can see the 2 in a. This means that the parabola is stretched by a factor of 2.


Now you might be wondering what stretched is, and how we use it. That will be explained later, when we learn to graph the vertex form.


Next we can look at the h value:

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The h is -1.


  • We remember that h is responsible for the horizontal translation. We also remember that the vertex will translate the opposite direction of the sign.
  • You can see that the h is negative. This means that it will translate to the right, and it will translate by 1 unit.



After that, we can look at the k.

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The k is 3.


  • The k is used to translate the graph vertically. This will move the graph up or down.
  • Remember, unlike the h, the k moves in the same direction of the sign.
  • Because the k is +3, the graph will move 3 units up.



Finally, we know what will happen to our graph. It is reflected vertically, stretched by a factor of 2, translated 1 unit right, and translated 3 units up.


  • Notice the last two points: translated 1 unit right, and translated 3 units up.
  • Earlier, when describing h and k, we said that the translation would start at (0,0).
  • If it moves 1 to the right (x-axis), and 3 up (y-axis), what would the coordinates of the vertex be in (x,y)?



The vertex would be (1,3). This is why this form is called vertex form, because the vertex is written in the equation. Look at y=-2(x-1)^2+3.


  • See -1? That's the 1 of the vertex (x-axis). Remember, the sign is opposite.
  • See +3? That's the 3 of the vertex (y-axis).



Try to make the equations for the following:


  1. Stretched by a factor of 2. Vertex: (-3,4)
  2. Reflected in the x-axis. Compressed by a factor of 0.5. Vertex: (2,-5)
  3. Reflected in the x-axis. Stretched by a factor of 4. Vertex: (1,1)


Write the properties for the following parabolas based on the equations. Include stretch/compression, whether or not it is reflected in the x-axis, the vertex along with the translations it went through to get there, and if the vertex is a minimum or maximum point.


  1. y=-3(x+2)^2-5
  2. y=0.5(x-5)^2+2
  3. y=2(x-1)^2-7



We are almost ready to graph using vertex form. All that is left is to learn of something called the step pattern:

The Step Pattern

Remember earlier we said that in a quadratic equation, the variable is squared. This is where that can be important. The step pattern is used to help graph using vertex form. It is used to find the points of a graph.


Take a look at the graph for y=x^2:

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This is a parabola, with no stretch, reflection or translation. This is a simple parabola: y=x^2. The step pattern for this is "Over 1, Up 1. Over 2, Up 4". Lets find out why this is the step pattern:
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As you can see, when x is 1, y is 1. But when x is 2, y is 4, and when x is 3, y is 9. What do you notice? The y is x^2, x squared.


This is what is meant by the variable is squared. The y is equal to x times x. So if x is 4, the y is 16. This is a step pattern. The step pattern is a way to describe what the y is when the x is a certain number, and in this case: 1 and 2. Remember, the step pattern for this parabola is "Over 1, Up 1. Over 2, Up 4".


You begin the step pattern at the vertex. You can use this step pattern moving both left and right of the vertex, because a parabola is symmetrical from the axis of symmetry. This means that you can go left 1, up 1, left 2, up 4 AND right 1, up 1, right 2, up 4. You use the step pattern to plot points on your graph so you can sketch the parabola.


Now you might be wondering "Okay, so I know how to graph y=x^2. Just start at (0,0), and go over 1, up 1, over 2, up 4. But how would I graph something like y=-2(x-1)^2+3?"


The step pattern will change as the a value changes. When the graph is stretched or compressed, the steps will increase or decrease, accordingly. Also, if a is negative, then the step pattern will be "Over, Down. Over, Down" instead of up.


To find a step pattern for an equation, you can use the basic step pattern of Over 1, Up 1, Over 2, Up 4. We can just multiply all of the y values by the a value. For our equation y=-2(x-1)^2+3, we will multiply the y values by -2. It will look like:


Over 1, up 1a. Over 2, up 4a.


Using this, the new pattern would be: Over 1, down 2. Over 2, down 8.


Do you notice how this changes? When the a value changes, the step pattern changes. To recap, just multiply the y value by the a value, and you found the new step pattern.


So now we have the vertex (because of the translations), the direction of opening (reflection), and the step pattern. we are ready to graph!

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Graphing With Vertex Form

Let's look at y=-2(x-1)^2+3, and how to graph it. Here are the steps:


1. Notice the stretch and the direction of opening.


In this equation, the vertical stretch is by a factor of 2. Also, the direction of opening is down because it is negative.


2. Find and plot the vertex.


We have already found the vertex for this equation. By looking at the equation, we know that the vertex is (1,3). We can plot this point on the graph.

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3. Find points on the graph.


You can do this by finding the step pattern of the graph. Remember, to find the step pattern by an equation, multiply the original step pattern by the a value. It is Over 1, up 1a, Over 2, up 4a. The new step pattern for this equation is "Over 1, down 2, Over 2, down 8".


Begin at the vertex that you plotted, and use the step pattern to find 2 points on both the right and left of the vertex. This means go left 1, down 2, left 2, down 8. Also go right 1, down 2, right 2, down 8.



Another way to do this is to sub in a point on the x-axis into the equation as the x, and solve for y. The x would be the x-coordinate of this point, and the y that you solved for would be the y-coordinate. For example, if our vertex is (1,3), I can look for the y when x is 2. so it would be:


y=-2(2-1)^2+3

y=1.


So a point on my graph would be (2,1). However, you should probably use the step pattern, as it is faster and you wont need to do the equation to find every point.

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4. Sketch the parabola.


Now that you have the vertex and 4 other points, you can draw the U-shaped parabola. Just sketch it like its connect the dots, but extend it a bit and add arrows to each side. Your graph should look something like:

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Let's do another example:
Vertex Form Graphing

Practice Word Problem

A soccer ball is kicked into the air, and follows the path of h=-0.25(t-6)^2+9. h represents height and t represents time.


1. Graph this equation.

2. For how long was the ball in the air?

3. What was the highest point the ball reached, in meters?

4. At how many seconds did the ball reach this height?



Because this is a practice problem, we will solve this for you, so you can take this as an example.


1. Graph this equation.

2. For how long was the ball in the air.


For this question, we can look at the 2 x-intercepts. The first one on the left would represent where the ball was kick. From there to the other x-intercept would be the time that the ball was in the air.

The ball was in the air for 12 seconds.


3. What was the highest point the ball reached, in meters?


Now we need to look at the vertex, as it is the maximum point in this graph. We can look for the y-coordinate of the vertex, and that will be our highest point.

The highest the ball reached was 9 meters.


4. At how many seconds did the ball reach this point?


For this, we can look at the x-coordinate of the vertex.

The ball reached 9 meters at 6 seconds.

Determining the Equation of a Parabola When Given the Vertex

You may come across some questions that ask you to find the equation of a parabola, and you are given the vertex and a point that the graph passes through. This may seem confusing, but here's how you do it:


Let's say the vertex is (-2,8), and the parabola passes through the point (2,0).


Remember how we found the vertex by looking at h and k? Also, remember how it was mentioned that the vertex was actually in the equation and that's why the form was called vertex form. Well, if you're given the vertex in a question, you already have 2 variables: h and k.


h is the x-coordinate of the vertex, but with the opposite sign. In this case, it would be 2.

k is the y-coordinate of the vertex. In this case, it would be 8


You can put these points into the vertex form equation. So at this point, our equation is:

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Next, we know that this parabola passes through the point (2,0). We can substitute these values in for x and y in our equation. Now, our equation is:

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At this point, the only variable left is a, meaning that we can solve for a by isolating it.

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Now that we have a, we can place it into our equation, and change x and y back to variables. Finally, our equation is:

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Remember, when you're solving one of these questions, first, put the h and k values into the equation, then substitute x and y to solve for an a value, and finally, but a into the equation to finish the equation.



Let's try another one.


"Find the equation for a parabola that was a vertex of (2,6) and passes through the point (5,3)."


The solution would look like:

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Therefore, the equation for the parabola is:

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FACTORED FORM

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What is Factored Form?

Factored form is a bit different than vertex form. Instead of graphing with the vertex and step pattern, we use the zeros, also known as the x-intercepts. Once you have the x-intercepts, you would use them to solve for the axis of symmetry. Finally, you would use the x-intercepts and the axis of symmetry to solve for the optimal value. Where the axis of symmetry and the optimal value cross is the vertex. After we have our vertex, we can plot it, along with the x-intercepts and sketch the parabola. There is only two parts that are identifiable directly from the equation in the factored form:

x-intercepts

As mentioned before, you graph factored form by using the x-intercepts. There are always 2 x-intercepts in a factored form equation. The x-intercepts are the r and s in the equation. Once again, the signs are flipped, similar to the h in vertex form. This means that if the x-intercept is negative, it will be positive when graphing, and vice versa. Here are what the x-intercepts look like:
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In this equation, the 2 x-intercepts are 5 and -3. Remember the signs are opposite.

a Value

You can also get the a value from factored form, at the beginning of the equation. Once again, the a value represents the stretch. However, the a value is not needed for graphing. You can still use it to make sketching the graph easier, or to check your graph by using the step pattern to see if you sketched it correctly.
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Finding the Vertex with x-intercepts

The first step of finding the vertex is to find the axis of symmetry. Finding the axis of symmetry is simple. Add the 2 x-intercepts, then divide by 2. It looks like:
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We can use the example above. We already know that the x-intercepts are 5 and -3. So we will add the 5 and -3, then divide it by 2.
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Now, we know that our axis of symmetry is 1. This is also the x-coordinate of the vertex. All that is left is to find the y-coordinate of the vertex. Now that there is only 1 variable left, we can just substitute the x-coordinate into the equation to find y.
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At this point, we also have the y-coordinate of the vertex, or the optimal value. We can put the x-coordinate and the y-coordinate together to get the vertex. The x-coordinate is 1, and the y-coordinate is -32. The vertex of the parabola is (1,-32). We have the vertex, and the 2 x-intercepts, meaning we're reading to graph!

Graphing with the x-intercepts and the Vertex

It is really simple now that we have our vertex and x-intercepts. Here are the steps.


1. Plot the x-intercepts. We will put points at (5,0) and (-3,0). Remember, for x-intercepts, the y-value is 0.

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2. Plot the vertex at (1,-32)
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3. Finally, sketch in the parabola. Connect the vertex to each of the x-intercepts, making u-shape.
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...and we're done. If you want, you can also check the parabola with the step pattern, as mentioned in vertex form. In this case, the parabola is "over 1, up 2, over 2, up 8."


Let's try another example:

Factored Form Graphing

STANDARD FORM

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What is standard form?

Standard form is the final of the 3 forms of quadratic equations, and is a trinomial. It is shown above. An example with real numbers would be:
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Graphing standard form

To graph standard form, we will need to factor it. Factoring is finding that is multiplied together to make the expression. If we were to factor standard form, we would get an expression in factored form, which we can then use to graph (because we learned how to do that above).


Factoring will be taught below, after multiplying binomials. However, here is an example of graphing standard form. Keep in mind what is happening so that you will understand how to graph.

Standard Form Graphing

MULTIPLYING BINOMIALS

What is a binomial?

A binomial is an algebraic expression with only 2 terms. An example is:
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Multiplying 2 binomials would be multiplying 2 of those expressions above. Multiplying binomials uses the distributive property, but with both terms. Multiplying binomials is also how you convert factored form to standard form, and vertex form to standard form. A question that involves multiplying binomials could look like:
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To multiply binomials, you want to make sure that both terms in the first binomial multiplies with both terms in the second. It is better explained like this:
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To multiply these 2 binomials, we will need to do the following:
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As you can see, the distributive property applies to both terms in both binomials. After we know what to do, we can multiply, and then add the like terms. Here is the solution:
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Now you may have noticed that the answer looks like standard form. That is an accurate observation. Multiplying binomials is how you can convert factored form to standard form. If we remember, factored form is 2 binomials multiplied together, and standard form is expanding that. For example:
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In this example, an equation in factored form turns into an equation into standard form. Multiplying binomials is how to convert factored form to standard form.


If you're wondering how to convert standard form to factored form, that is done by factoring, which is further down on this site.

Perfect Square Binomials

These special binomials are perfect squares, meaning all of the binomial is squared. This is also what you would use to convert vertex form to standard form. Examples are shown below:

Positive

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If we were to expand this, it would look like:
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If we were to substitute and complete this expression with our binomial, we would get:
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Negative

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This can be expanded to:
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If we were to substitute and complete, we would get:
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Product of sum and difference

This special case is multiplying a sum and difference binomial, as shown below.
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In this special product, there is a positive b and a negative b. The equation also changes in this scenario as well. The equation for this special product is:
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The reasoning for this is because the 2ab is cancelled out because 1 is positive and one is negative. This can be shown with the distributive property:
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An example solution of this would look like:
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Algebra tiles

You can also use algebra tiles to multiply binomialsAlgebra tiles are visual representations of algebraic expressions. To use algebra tiles, multiply what we have on the side by what we have on the top or bottom. For example, look at this:
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Here, we have a 1 on the side, and a 1 on the bottom. the expression for this would be 1*1. We have one 1 on the side, and one at the bottom. We multiply these to get 1 in the middle, which is the answer. There are 3 important blocks to know:
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When 1 is multiplied by 1, we get a 1 in the middle.
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When 1 is multiplied by x, we get an x in the middle. This is because 1 times x is x.
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When x is multiplied by x, we get x^2 in the middle. This is because x times x is x squared.
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With algebra tiles, you always the the middle to be a full rectangle. This means that the sides don't have missing tiles. Let's try an example.
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First, let's put the (x+2) on the side.
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Next, let's put the (x-5) on the bottom.
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Finally, we want to multiply the side by the bottom, and fill in the middle.
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Notice the black outline. As you can see, it is a perfect rectangle, with no parts missing. Now, we can count the blocks to get the answer. There is one x^2, five -x, two positive x, and ten -1. We can add these together to get our final answer:
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FACTORING

What is factoring?

Factors are numbers you can multiply together to get another Number. Factoring finding out what those numbers are. Factoring will also help us get from standard form to factored form, and even standard form to vertex form (completing the square). We are going to learn how to factor algebraic expressions in a few ways. We will cover:



  • Common factoring
  • Factoring by grouping
  • Factoring simple trinomials
  • Factoring complex trinomials
  • Special products
  • Factoring perfect square trinomials

Common factoring

Let's start factoring with the easiest part: common factoring. Common factoring is simply finding something that is a factor of all the terms, and dividing everything by it. To clarify, let's look at an easy example:
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What is a number in the expression above that is a factor of all the terms (can be divided by everything). The common factor would be 2. So now, we want to divide everything by 2.
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Finally, we want to place the common factor outside by using brackets. This way, it is the factor times what is inside the brackets.
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The reason why we want to do this is because now, both this expression and the original example are equal to the same thing. With common factoring, we should be able to use the distributive property and multiply, and the end result should be what we started with. This is a method to check our answer.


Now, let's try a harder example:

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In this example, there isn't the same number in all the terms. This means that we will need to find something common in all of them. In this case, 3 is the highest common factor. We can divide everything by 3.
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Finally, we can close the brackets and multiply it all by the common factor: 3.
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Factoring by grouping

Factoring by grouping is splitting a long expression up into two parts, finding a common factor for each of the two parts, and then finding a common factor for the whole expression. This will make more sense when it is shown with an example:
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As you can see, there is not a common factor that can be divided by all four terms. However, there are common factors if we split this up.
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We are now working with two different common factors. The common factors are much more clear now. At this point, we can use what we learned from the "Common factoring" section above. We will need to get this to the point where the factor is outside of the bracket.
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Now that we have factored each group, what do you notice. Look at what is inside the brackets. Both brackets have the same thing: x + y. This means that we can factor this expression further. We can divide each group by (x+y).
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The x+y multiplication and division will cancel each other out, and we will have m+2 remaining. However, we will need to put the m+2 in brackets, and the factor (x+y) outside of it.
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This is the end result, and the expression can not be factored any further. If we were to multiply the (x+y) and (m+2), we would get what we had at the start: mx + my + 2x + 2y.

Factoring simple trinomials

Remember standard form? It was a trinomial. Standard form looked like:
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In a simple trinomial, a = 1. An example of a simple trinomial is:
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We want to factor the simple trinomial into factored form. Let's look back at what that looks like:
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There is a relationship between factored form and standard form. We already know that when you multiply the 2 binomials of factored form, you get standard form. Now, we are trying to get standard form into factored form. Here are some important things to remember:


  • The b value of standard form is equal to r + s of factored form.
  • The c value of standard form is equal to r x s of factored form.

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This means that to factor a simple trinomial, we need to find the factors of c, but there is a catch: we need 2 numbers that have a product of c but a sum of b. In other words, the numbers need to add to b and multiply to c.


To do this, we need to find factors of c, and whichever pair equals to b is the factors we are looking for. Lets look back at our example.

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We need to find the factors of 2, and it must add to 3. In this case, it is easy because 2 only has 2 factors: 1 and 2. If we add 1 and 2, we will get 3. However, there is also a chart you can use to get the factors (which can be helpful in harder questions. It looks like:
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In this chart, you can write the factors of c, 2 in this case, in the first column. Next, you will multiply them to ensure that they equal c. Finally, you will add them to check if they equal b. If they equal b, these are the factors you are looking for. If they do not equal b, you can check the next factors of c.


So now we have the numbers that have a product of 2 and a sum of 3. We can finally fully factor.


Since we already know that the factors of x^2 are x and x, we can place those.

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Next, we just need to add the factors to the expression. One factor will be in the first bracket, and the other will be in the second bracket.
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And that is the factored form of the expression. Factoring simple trinomials is simple, as the a value is always 1. This let's us know that there is x with no coefficient in each bracket. Let's try one more, but with a negative number:
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Here, we need numbers that when multiplied equal 4, and when added equal -5. We know that both the numbers must be negative, because the b is negative and the c is positive. If we remember integers, negative times negative equals positive. So let's use the chart to find our factors.
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Notice the sum of 1 and 4. It is 5. If you notice that you find the answer, but with the wrong sign, you can flip the sign of your factors to make the sum negative.
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And we found our factors. Now we just need to insert it into the factored form, and the factored form of this equation is:
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Factoring complex trinomials

Factoring complex trinomials is different from simple trinomials. Sure, they both follow standard form. However, in a complex trinomial, a does not equal 1. Also, when factoring complex trinomials, you also need to find the factors of the first term. A complex trinomial can look like this:
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Factoring complex trinomials can take time, and very often you will need to use trial and error. Once again, you want to try to make it into factored form. Here are the steps to factor a complex trinomial:


  1. First look for common factors, and remove them. If possible, turn the complex trinomial into a simple trinomial by factoring the a value from the trinomial. If you can do this, then factor it like a simple trinomial.
  2. Write the factors of the first and last term. In a simple trinomial, you only need the factors of the 3rd term, because the factors of the first term were always x and x. However, in a complex trinomial, the a value is not 1, so you must find the factors.


Once again, you will need specific factors. Each of the factors of the first term must multiply with one of the factors of the third term. Then, when you add it all up, it should equal the middle term.

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Let's try an example. For the example at the beginning, we can make it into a simple trinomial by factoring the 3.
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Let's try a harder trinomial. One where we will need to find the factors of the first and last term. Let's try this:
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For this example, there are no common factors. We need to find the factors of the first and last term. We can use a chart for this too:
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We can use this chart. We will need to put the factors in each box, and then cross multiply. Once we have that, we will add it to find the total. The total must equal bx, or -15x in this case. If it does not equal that, then we will need to try again but with different factors.
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As you can see, the total did not equal -15x. So we must try again with different factors.
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This time, the total did equal -15x. This means that we have our factors. All that's left is to put it into factored form. We can just take one row and put it together, as shown below.
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Factoring the Difference of Squares

An example of difference of squares can look like:
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Now, the name is difference of squares. Sure, the x is squared, but is that it? Nope. 100 is also a square. It is the square of 10. So in the difference of squares, we have a square subtract a square.


Look back to MULTIPLYING BINOMIALS, and look at The Product of Sum and Difference. Remember:

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(a+b)(a-b) turns into a squared subtract b squared. Now look at this expression once again:
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Remember, the 100 is also a square. What do you notice?


This expression follows the the a squares subtract b squared expression. In this case, x is a and 100 is b. This must mean that it is also equal to (a+b)(a-b). If we want to factor this expression, we must get it back to (a+b)(a-b).


Another reason why this equation must be (a+b)(a-b) is because there are 2 squares, and the sign is negative. A sign becomes negative when it is positive and negative, meaning that one must be positive and the other must be negative.

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In order to find a and b, we must find the square root of both of these.
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Now that we know our a and b, we can fill it into (a+b)(a-b).
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And we have our factors. We can check this by expanding and simplifying it.
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Let's try another example:
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In this expression, 25 is squared, x is squared and 64 is squared. If we find the square roots of all of these, they are:

5

x

8

This means a is 5x and b is 8. We can fill this into (a+b)(a-b).

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Finally, we can check this by expanding and simplifying:
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Factoring Perfect Square Tinomials

Remember this equation from perfect squares:
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This is a perfect square trinomial. Both a and b are squared. We will now see how to factor this.


To factor this, you can just take a and b, put it in a bracket and square it. The factored form of this looks like:

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Sounds very simple, right. However, sometimes you need to identify if the numbers have already been squared. For example, let's use this expression:
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You might be thinking, "Hey! That doesn't look like anything above!" Well actually, it is. Remember in the difference of squares, the numbers were already squared? That is what's going on here. the 25 and 36 are already squared. So let's find the square roots.
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Not it is starting to look like the algebraic expression, right? All that's left is to find the middle. 22x has to be equal to 2ab. We already know that a is 5 and b is 6. If we do 2(5)(6), we get 60. This just confirms that the equations are equal.
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Now that we have our equation like this, it looks just like
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All that is left is to put the substitute the numbers from the first expression into the last. Our factored form would be:
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We can check it over by expanding it or using the perfect square expression.
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Let's try a new example:
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COMPLETING THE SQUARE

Put it together

We know the three forms of quadratic equations, and all that's left is to put it together and link the start to the end. Completing the square is a way to turn standard form into vertex form. Briefly, The steps for completing the square are:


1. Make the standard form equation look similar to vertex form.


This can be done by factoring our the a value and separating the c to make it look like k from vertex form.


2. Expand what's in the brackets to make it a perfect square trinomial.


You will need to add and subtract the c from the perfect square trinomial.


3. Factor the perfect square trinomial.


4. Apply the distributive property from a.


5. Add like terms.


This will be better shown with an example:

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Step 1: Make it look similar to standard form.


In this step, we will factor our the a value, and separate the c by brackets.

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Step 2: Make a perfect square trinomial.


In this step, we need to turn what is in the middle into a perfect square trinomial. To do so, we need a c value. In order to get the c value we need for a perfect square trinomial, divide the b value by 2. (This is because in a perfect square trinomial, the middle value equals to 2ab from the binomial.) In this case, -4 divided by 2 is -2. -2^2 is our c value, because a perfect square trinomial is a^2 + 2ab + b^2.


We will need to add the c value, and then subtract it. This is because we cannot just add a number into the equation, as that changes the equation completely. However, + c - c = 0, so we can do that.


In the final part of this step, the perfect square trinomial is in a bracket in a bracket. This is only to help organize, and is not necessary. Everything inside the bracket will be multiplied by the distributive property of 3 soon, however, putting the perfect square trinomial in a bracket just helps you see what will be factored.

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Step 3: Factor the perfect square trinomial.


Now we will factor the perfect square trinomial. We need to find the square roots of x^2 and 4, which is x and 2. However, the b term is negative, meaning that the 2 is negative. Our factored form of the perfect square trinomial is (x-2)^. We can keep this inside of the bracket, as it will still be multiplied by 3.


The -4 (which was used to counter the +4 from the perfect square trinomial) will remain inside the bracket, and wasn't factored because it wasn't a part of the perfect square trinomial. It will also be multiplied by the 3.

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Step 4: Apply the distributive property.


Next, we will multiply everything that is inside of the bracket by the a value, which is 3. The (x-2)^2 will be multiplied by the 3, and the -4 will be multiplied by the 3 as well.

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Step 5: Add like terms.


Finally, we can finish this by adding like terms. We must the number we left outside of the bracket to the number that was multiplied by the a value. In this case, the number outside was -5, and the number we got after multiplying by 3 was -12.

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And now, the equation is in standard form.


The a is 3, the h is -2, and the k is -17.


Let's try another example. This example will have a negative a value, so we will need to factor out -1 from everything inside the brackets.

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SOLVING QUADRATIC EQUATIONS

Solving a quadratic equation means getting the roots/x-intercepts. When you are solving a quadratic equation, you are solving for x when y is 0. By solving a quadratic equation, we can find the axis of symmetry, optimal value, and all together, the vertex. This will allow us to graph our equation.

Solving Quadratic Equations by Factoring

One way to solve quadratic equation is to factor. To use this method, we must get the equation into factored form. If it is in standard form, factor it.


In factored form, if the equation is equal to 0 (the y), that means that one or both of the factors is equal to 0. So solve for x, we will set each of the factors to 0, and isolate x. It should look like this:

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The 2 numbers that equal to the x's are the roots. This is the same thing that you do to graph form factored form, and is very simple. Let's try an example:
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The roots are 6 and -2. This is the solution of the quadratic equation.


Next, let's try an equation with a coefficient in front of x. Since the x is multiplied by the coefficient, the only new step is to divide both sides by the coefficient. Let's try an example:

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As you can see, the only other steps were to divide the coefficient to isolate x. The solutions to this quadratic equation were -4 and 4.

Solving Quadratic Equations Using the Quadratic Formula

The quadratic formula is another way to find the solutions of a quadratic equation. The quadratic equation can be used directly form standard form. It looks like this:
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The variables in the formula will be substituted with the variables from standard form. Remember, standard form looks like:
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The a from standard form will replace the a in the formula. The same thing with the b and c.


Also, look back at the formula. There is a plus and minus sign together. This means that the formula can be used twice: one with a plus in that spot and one with a minus in that spot. Since the formula can be used twice, each time will give a different root, for a total of 2 roots.


Let's try an example:

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We will put the values of a, b, and c into the quadratic formula so that we can solve the 2 roots.
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Now, we've reached the plus-minus sign. There are 2 possible ways to go from here. The first is to do the equation with the plus. That will give us 1 root. The other is to do the equation with a minus. That will give us the other root. We must do both ways.
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These are the two roots: 0.87 and -1.54. That is the solution to the quadratic equation.


There is one more thing to bring to attention. Look back to the work for the solution. There is a square root in there. The number that is being square rooted is called a discriminant. In this quadratic equation, the discriminant was 52. The discriminant can help us know how many roots there will be.


  • If the discriminant is positive, there will be 2 roots. This is because the number will be squared, and either added to or subtracted from the number before it.
  • If the discriminant is negative, there will be no roots. This is because a negative number cannot be square rooted, as a square is a number times itself. To get a negative sign, the signs multiplied must be opposite, so you cannot square root a negative following this concept.
  • If the discriminant is zero, there will be 1 root. This is because the square root of 0 is 0, so nothing will be added or subtracted to the number before it, and it will just be the number -b divided by 2(a).

Solving Quadratic Equations by Isolating X

The final way to solve a quadratic equation is by isolating x. This way can be used in vertex form, and is very simple. It is isolating the variable to find out what it is. This method also includes a plus-minus sign. This way is better shown with an example, as there is no real formula. Let's use this example:
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We will start isolating the x.
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As you can see, the plus-minus sign is on the other side than the x. This can give us 2 roots: 1 when the sign is positive, and the other when the sign is negative.


Like the quadratic formula and the discriminant, there is also a way to find out how many roots we will have by looking at the equation. For vertex form, we must look at the a and k values:


  • If a is positive, and k is positive, there will be 0 roots.
  • If a is positive and k is negative, there will be 2 roots.


  • If a is negative and k is positive, there will be 2 roots.
  • If a is negative and k is negative, there will be 0 roots.


  • If k is zero, there will be 1 root regardless of what the a is.

LINKING TOPICS

Although I went through this throughout the website, I haven't said the links explicitly, so here are some important links to know:



  • Expanding binomials will help you get from factored and vertex form to standard form.
  • Completing the square will get you from standard form to vertex form. This also requires factoring (perfect square trinomials).
  • Factoring will also help you get from standard form to factored form.
  • Factoring relates to graphing because it can help us get roots.
  • We can use quadratic formula from standard form to get roots.
  • Roots can help us get the vertex, and graph the parabola.
  • Vertex form relates to graphing because it shows us the vertex and translations, and also the step pattern.
  • Learning about discriminants will help us know how many roots there are.
  • Algebra tiles can help with multiplying binomials and factoring.

WORD PROBLEMS

We can go over some practice word problems. We will do a few that will display how word problems that are included in quadratics can look like.

Motion Problems

1. A tennis ball is thrown at an initial velocity of 9.2 m/s, from a height of 1.6 m above the ground. The height of the tennis ball, in meters, above the ground after t seconds is modelled by this equation:
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a) How long does it take for the tennis ball to fall to the ground, rounded to the nearest tenth of a second?


b) Find the times when the tennis ball is at a height of 4.5 m above the ground. Round your answers to the nearest tenth of a second.


c) What is the maximum height of the tennis ball? At what time does it reach this height? Round your answers to the nearest tenth.



To answer these:

For a), we need to find the roots. For b), we need to find the roots when y=4.5. For c), we need to find the vertex. We can answer these questions using the quadratic formula.

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Revenue Problems

Calculators are sold to students for 20 dollars each. 300 students are willing to buy them at that price. For every 5 dollar increase in price, there are 30 fewer students willing to buy the calculator. What selling price will produce the maximum revenue and what will the maximum revenue be?


To answer this:

We will first need to define our variables. Next, we will need to make our equation. We can use factored form for this. Then, we must find our roots and vertex.

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Area Problems

1. The width of a rectangle is 12 metres less than the length. The area of the rectangle is 320 square metres.


a) Write an equation that can be used to find the length and width of the rectangle.


b) What are the dimensions of the rectangle?


To answer this:

First, define your variables. Next, create an equation (remember, area = l*w.) Then, solve for your variables. Here's an important tip: dimensions cannot be negative. This narrows your roots to only one. Also, drawing a diagram can help.

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Consecutive Number Problems

The product of two consecutive numbers is 1482. What are the two numbers?


To answer this:

First, define your variables. Next, create an equation. After this, you can solve the quadratic equation.

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Right Triangle Problems (Pythagoras' Theorem)

One leg of a right triangle is 1 cm longer than the other leg. The length of the hypotenuse is 9 cm greater than that of the shorter leg. Find the lengths of the three sides.


To answer this:

Define variables. Use Pythagoras' Theorem. Base the lengths off of 1 variable. Solve for that variable. (Hint: a diagram can be helpful.)

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Optimization Problem (Area)

You have a 500-foot roll of fencing and a large field. You want to construct a rectangular playground area. One side will not need fencing because it is a wall of a building. What are the dimensions of the largest such yard? What is the largest area?


To solve this:

First, define variables and create an equation. Next, you must find roots, axis of symmetry and the optimal value. Also, a diagram might help.

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REFLECTION OF THE UNIT

I believe that quadratics was an interesting unit. It was different than what we did in grade 9 - linear systems. Quadratics included lines that curved, called parabolas, due to a squared variable. Quadratics is a bit more challenging than linear systems, as there was also a lot more to cover in a shorter amount of time. In this unit, we learned the 3 types of quadratic equations, multiplying binomials, factoring, algebra tiles, how to change between the 3 equations, word problems and more. Quadratic word problems can be useful in real life, such as optimization problems. It could be useful to find a playground that will give the largest area, or the price of something that will bring in the most revenue. Many of these things can be used in real life jobs, including architecture, business, etc.


Here is one of my assessments from this unit:

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This assessment was the mini-test 2. It included multiplying binomials, factoring, graphing, and a communication problem.



  • For the matching part, I was able to identify the types of equations and which methods would be used from practice.




  • Number 6 was multiplying binomials. I multiplied the 2 binomials, then added like terms.




  • Number 7 was factoring. The first question was a complex trinomial. I found the factors by finding the factors of a and c, and making sure that when multiplied and added, they would add to b.




  • Number 8 was factoring, and then solving for the roots. To do this, i factored each equation, then made it equal to 0. After this, I isolated the x's to find the 2 roots.




  • In number 9, I once again used factoring to find the roots. Once I had the roots, I found the axis of symmetry and the optimal value, which gave me the vertex. Finally, I graphed it by plotting the vertex and the roots.




  • In number 10, I created an equation for each shape. I made an equation to find the area of the square and an equation to find the area of the rectangle. I then subtracted the area of the rectangle from the area of the square, and added like terms. This gave me the area of the shaded region. I then factored it in the next question using common factoring.




  • In number 11/12, I was able to find the answer that would give one root, which was 30. I got this because the trinomial in the question was a perfect square trinomial. When there is a perfect square trinomial, there is only 1 root, so I solved for the b of the equation.



All in all, this assessment was good. I got a great mark, and I learned a lot from this part. Factoring and multiplying binomials is an important concept to understand, and I'm proud that I did good on the test.



That's all that's left on this Smore. However, if there is something you need clarification on, or need more practice, you can visit: