### Compiled by: Sabrina Singh

The prefix quad- means “four” and quadratic expressions are ones that involve powers of x up to the second power (not the fourth power). So why are quadratic equations associated with the number four?

Within this unit we moved on from linear relations to quadratics relations. We touched upon the many components of quadratics which ranged from a basic introduction to graphing equations in a variety of forms. Below is the structure of the lessons and activities:

Introduction

• Properties Of A Parabola
• Transformations

Types of Equations

• Factored Form a(x-r)(x-s)
• Vertex Form a(x-h)²+k
• Standard Form ax²+bx+c

Vertex form

• Finding an equation when only given the vertex
• Graphing vertex form

Factored Form

• Factors and zeroes
• Polynomials (factored form to standard form)
• Common factoring
• Factoring simple trinomials
• Factoring complex trinomials
-Special cases
- Difference of squares
• Perfect square
• Factoring by grouping
• Graphing factored form

Standard form

• Completing the square (standard to vertex)
• Graphing from standard form

Word Problems

• Motion
• Revenue
• Numeric

## Introduction

An expression of the form ax²+ bx+ c with x the variable and a, b, and c fixed values (with a ≠ 0 ) is called a quadratic. To solve a quadratic equation means to solve an equation that can be written in the form ax²+ bx+ c=0 .

Explain the meaning of the term function, and distinguish a function from a relation that is not a function, through investigation of linear and quadratic relations using a variety of representations (i.e., tables of values, mapping diagrams, graphs, function machines, equations) and strategies.

• Substitute into and evaluate linear and quadratic functions represented using function notation, including functions arising from real-world applications.

• Explain the meanings of the terms domain and range, through investigations using numeric, graphical, and algebraic representations of linear and quadratic functions, and describe the domain and range of a function appropriately.

• Explain any restrictions on the domain and the range of a quadratic function in contexts arising from real-world applications.

• Determine, through investigation using technology, the roles of a, h, and k in quadratic functions of the form f(x)=a(x – h) 2 + k, and describe these roles in terms of transformations on the through investigation with and without technology,from primary sources, using a variety of tools, or from secondary sources, and graph the data.

• Determine, through investigation using a variety of strategies, the equation of the quadratic function that best models a suitable data set graphed on a scatter plot, and compare this equation to the equation of a curve of best fit generated with technology.

• Solve problems arising from real-world applications, given the algebraic representation of a quadratic function.

While the first derivative can tell us if the function is increasing or decreasing, the second derivative tells us if the first derivative is increasing or decreasing. If the second derivative is positive, then the first derivative is increasing, so that the slope of the tangent line to the function is increasing as x increases. We see this phenomenon graphically as the curve of the graph being concave up, that is, shaped like a parabola open upward. Likewise, if the second derivative is negative, then the first derivative is decreasing, so that the slope of the tangent line to the function is decreasing as x increases. Graphically, we see this as the curve of the graph being concave down, that is, shaped like a parabola open downward. At the points where the second derivative is zero, we do not learn anything about the shape of the graph: it may be concave up or concave down, or it may be changing from concave up to concave down or changing from concave down to concave up
This table indicates that the equation y=x² is a quadratic relation because the first differences are not equal while the second differences are.

## Parts Of Parabola

The Vertex of a Parabola

• The vertex of a parabola is the point where the parabola crosses its axis of symmetry. If the coefficient of the x^2 term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the “U”-shape. If the coefficient of the x^2 term is negative, the vertex will be the highest point on the graph, the point at the top of the “U”-shape.

The Axis of Symmetry

• The axis of symmetry of a parabola is the vertical line through the vertex. For a parabola in standard form, y =ax^2 + bx + c, the axis of symmetry has the equation.Note that –b/2a is also the x-coordinate of the vertex of the parabola.

Intercepts

• You can find the y-intercept of a parabola simply by entering 0 for x. If the equation is in standard form, then you can just take c as the y-intercept. For instance, in the above example: y =2(0)2 + (0) – 1=–1 So the y-intercept is –1.
• The x-intercepts are a bit trickier. You can use factoring, or completing the square, or the quadratic formula to find these (if they exist!).

## Graphical Transformations of Functions

In this section you will learn how to draw the graph of the quadratic function defined by the equation f(x) = a(x − h) 2 + k. You will quickly learn that the graph of the quadratic function is shaped like a "U" and is called a parabola. The form of the quadratic function in equation is called vertex form, so named because the form easily reveals the vertex or “turning point” of the parabola. Each of the constants in the vertex form of the quadratic function plays a role. As you will soon see, the constant a controls the scaling (stretching or compressing of the parabola), the constant h controls a horizontal shift and placement of the axis of symmetry, and the constant k controls the vertical shift. Let’s begin by looking at the scaling of the quadratic.

The graph of the basic quadratic function f(x) = x 2 is called a parabola. We say that the parabola “opens upward.” The point at (0, 0), the “turning point” of the parabola, is called the vertex of the parabola. We've tabulated a few points for reference in the table and then superimposed these points on the graph of f(x) = x 2.

Now that we know the basic shape of the parabola determined by f(x) = x 2 , let’s see what happens when we scale the graph of f(x) = x 2 in the vertical direction. For example, let’s investigate the graph of g(x) = 2x 2 . The factor of 2 has a doubling effect. Note that each of the function values of g is twice the corresponding function value of f in the table
When the points in the table are added to the coordinate system , the resulting graph of g is stretched by a factor of two in the vertical direction. It’s as if we had put the original graph of f on a sheet of rubber graph paper, grabbed the top and bottom edges of the sheet, and then pulled each edge in the vertical direction to stretch the graph of f by a factor of two. Consequently, the graph of g(x) = 2x 2 appears somewhat narrower in appearance, as seen in comparison to the graph of f(x) = x 2 . Note, however, that the vertex at the origin is unaffected by this scaling. In like manner, to draw the graph of h(x) = 3x 2 , take the graph of f(x) = x 2 and stretch the graph by a factor of three, tripling the y-v

## Vertical Translations

The graph of f(x) = x ^2 + 1 is shifted one unit upward from the graph of g(x) = x 2 . This is easy to see as both equations use the same x-values in the table , but the function values of g(x) = x^ 2 + 1 are one unit larger than the corresponding function values of f(x) = x 2 .

Note that the vertex of the graph of g(x) = x 2 + 1 has also shifted upward 1 unit and is now located at the point (0, 1).

## Horizontal Translations

The graph of g(x) = (x + 1) shows a basic parabola that is shifted one unit to the left. Examine the table and note that the equation g(x) = (x + 1)2 produces the same y-values as does the equation f(x) = x 2 , the only difference being that these y-values are calculated at x-values that are one unit less than those used for f(x) = x 2 . Consequently, the graph of g(x) = (x + 1)2 must shift one unit to the left of the graph of f(x) = x 2 , as is evidenced. Note that this result is counter intuitive. One would think that replacing x with x + 1 would shift the graph one unit to the right, but the shift actually occurs in the opposite direction. Finally, note that this time the vertex of the parabola has shifted 1 unit to the left and is now located at the point (−1, 0). We are led to the following conclusion.

A similar thing happens when you replace x with x − 1, only this time the graph is shifted one unit to the right.

## Vertical Reflections

Let’s consider the graph of g(x) = ax^2 , when a < 0. For example, consider the graphs of g(x) = −x^ 2 and h(x) = (−1/2)x^ 2

When the table is compared with the other table , it is easy to see that the numbers in the last two columns are the same, but they've been negated. The result is easy to see. The graphs have been reflected across the x-axis. Each of the parabolas now “opens downward.” However, it is encouraging to see that the scaling role of the constant a in g(x) = ax^2 has not changed. In the case of h(x) = (−1/2)x^2 , the y-values are still “compressed” by a factor of two, but the minus sign negates these values, causing the graph to reflect across the x-axis. Thus, for example, one would think that the graph of y = −2x ^2 would be stretched by a factor of two, then reflected across the x-axis. Indeed, this is correct, and this discussion leads to the following property

• Vertex form: a(x-h)²+k
• Factored form : a(x-r)(x-s)
• Standard form: ax²+bx+c

## Vertex Form of Quadratic Functions

The vertex form of a quadratic function is given by f (x) = a(x - h)2 + k, where (h, k) is the vertex of the parabola.

When written in "vertex form":
(h, k) is the vertex of the parabola, and x = h is the axis of symmetry.
• the h represents a horizontal shift (how far left, or right, the graph has shifted from x = 0).

• the k represents a vertical shift (how far up, or down, the graph has shifted from y = 0).

• notice that the h value is subtracted in this form, and that the k value is added.

If the equation is y = 2(x - 1)2 + 5, the value of h is 1, and k is 5.

If the equation is y = 3(x + 4)2 - 6, the value of h is -4, and k is -6.

## Finding an equation when only given the vertex

Many real world situations that model quadratic functions are data driven. What happens when you are not given the equation of a quadratic function, but instead you need to find one? In order to obtain the equation of a quadratic function, some information must be given. Significant data points, when plotted, may suggest a quadratic relationship, but must be manipulated algebraically to obtain an equation.

3.4 Finding the Equation given Vertex

• When do I use each form?
• When you are given the vertex and at least one point of the parabola, you generally use the vertex form.
• When you are given points that lie along the parabola, you generally use the general form.

Let's use a vertex that you are familiar with: (0,0).

Use the following steps to write the equation of the quadratic function that contains the vertex (0,0) and the point (2,4).

1. Plug in the vertex.

2. Simplify, if necessary.

3. Plug in x & y coordinates of the point given.

4. Solve for "a."

5. Now substitute "a" and the vertex into the vertex form.

Our final equation looks like this:

f(x)= x^2

## Graphing Quadratic Function In Vertex Form

Vertex form of a quadratic function: f(x)= a(x-h)^2+k

• The graph f(x)= a(x-h)^2+k is the parabola y=x^2 translated an h united horizontally and k units vertically.

• y = a(x - h)2 + k ( y = 3(x - 2)2 - 4)
• Pull out the values for h and k.
If necessary, rewrite the function so you can clearly see the h and k values.
(h, k) is the vertex of the parabola.
Plot the vertex. y = 3(x - 2)2 + (-4)

h = 2; k = -4
Vertex: (2, -4)

3.2 Graphing from Vertex Form

Find two or three points on one side of the axis of symmetry, by substituting your chosen x-values into the equation.

For this problem, we chose (to the left of the axis of symmetry):

x = 1; y = 3(1 - 2)2 - 4 = -1

x = 0; y = 3(0 - 2)2 - 4 = 8

Plot (1, -1) and (0,8)

Plot the mirror images of these points across the axis of symmetry, or plot new points on the right side. Draw the parabola. Remember, when drawing the parabola to avoid "connecting the dots" with straight line segments. A parabola is curved, not straight, as its slope is not constant.

## Factored Form

An essential skill in many applications is the ability to factorise quadratic expressions. In this unit you will see that this can be thought of as reversing the process used to ‘remove’ or ‘multiply-out’ brackets from an expression. You will see a number of worked examples followed by a discussion of special cases which occur frequently with which you must become familiar. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that all this becomes second nature, and you eventually carry out the process of factorising simply by inspection. To help you to achieve this, the unit includes a substantial number of such exercises.

## Finding polynomial functions ( factors and zeros)

As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned addition and then you learned subtraction or first you learned multiplication and then you learned division. Each operation being the opposite of the other. Yet another example is one in which you learned the properties of exponents and then you learned about radicals or rational exponents.

Well, finding polynomials is the reverse of finding factors. In the previous lesson, you were given a polynomial and asked to find its factors and zeros. In this lesson, you will be given factors or zeros and asked to find the polynomial of lowest degree with real coefficients.

Example 1: Given the factors ( x - 3) 2 (2x + 5), find the polynomial of lowest degree with real coefficients.Let's analyze what we already know. ( x - 3) 2 is a repeated factor, thus the zero 3 has multiplicity of two. Did you know that the linear factorization theorem states that a polynomial of degree n has precisely n linear factors. And since we h been given three linear factors, the lowest degree of this polynomials must be three.

STEPS

• Rewrite the factors to show the repeating factor. ( x - 3)( x - 3)(2x + 5)
• Write factors in polynomial notation. The a is the leading coefficient of the polynomial P(x) = a( x - 3)( x - 3)(2x + 5)
• Multiply ( x - 3)( x - 3)
• Then multiply ( x 2 - 6x + 9)(2x + 5)
• For right now let an = 1
• Thus, the polynomial of lowest degree with real coefficients, and factors ( x - 3)2 (2x + 5) is: P(x) = 2x 3 - 7x 2 + 8x + 45.

The main concept is simply to multiply the factors in order to determine the polynomial of lowest degree with real coefficients.

## Multiplying-out brackets, and quadratic expressions

In this unit you will learn how many quadratic expressions can be factorised. Essentially, this is the reverse process of removing brackets from expressions such as (x + 2)(x + 3).

You will be familiar already with the well-known process of multiplying-out brackets. For example, you will have seen expressions like (x + 2)(x + 3) and then expanded these terms to arrive at a quadratic expression. Let us see how this works. The arrows in the figure below show how each term in the first pair of brackets multiplies each term in the second pair.

Then the like-terms, 3x and 2x, are collected together to give 5x. The result of multiplying-out the brackets is the quadratic expression x^2 + 5x + 6.
Like many processes in mathematics, it is useful to be able to go the other way. That is, starting with the quadratic expression x^2+5x+6, can we carry out a process which will result in the form (x + 2)(x + 3) ? The answer is: yes we can! This process is called factorising the quadratic expression. This would help, for example, if we wanted to solve a quadratic equation. Such an equation is formed when we set a quadratic expression equal to zero, as in x^2 + 5x +6=0 This is because the equation can then be written (x + 2)(x + 3) = 0 and if we have two expressions multiplied together resulting in zero, then one or both of these must be zero. So, either x + 2 = 0, or x + 3 = 0, from which we can conclude that x = −2, or x = −3. We have found the solutions of the quadratic equation x^2 + 5x + 6 = 0.

So, the ability to factorise a quadratic is a useful basic skill, which you will learn about in this unit. However, be warned, not all quadratics will factorise, but a lot do and so this is a process you have got to get to know!

To learn how to factorize let us study again the removal of brackets from (x + 3)(x + 2). (x + 3)(x + 2) = x^2 + 2x + 3x +6= x2 + 5x + 6

Clearly the number 6 in the final answer comes from multiplying the numbers 3 and 2 in the brackets. This is an important observation. The term 5x comes from adding the terms 2x and 3x. So, if we were to begin with x^2 + 5x + 6 and we were going to reverse the process we need to look for two numbers which multiply to give 6 and add to give 5.

? × ?=6

?+? = 5

What are these numbers ? Well, we know that they are 3 and 2, and you will learn with practice to find these simply by inspection. Using the two numbers which add to give 5 we split the 5x term into 3x and 2x. We can set the calculation out as follows.

x^2 + 5x +6

= x2 + 3x + 2x + 6

= x(x + 3) + 2x + 6 by factorizing the first two terms

= x(x + 3) + 2(x + 3) by factorizing the last two terms

= (x + 3)(x + 2) by noting the common factor of x + 3

The quadratic has been factorized. Note that you should never get these wrong, because the answer can always be checked by multiplying-out the brackets again!

Example 1:

Suppose we want to factorize the quadratic expression x^2 − 7x + 12. Starting as before we look for two numbers which multiply together to give 12 and add together to give −7. Think about this for a minute and you will realize that the two numbers we seek are −3 and −4 because

−4 × −3 = 12, and − 4 + −3 = −7

So, using the two numbers which add to give −7 we split the −7x term into −4x and −3x. We set the calculation out like this:

x^2 − 7x + 12 = x2 − 4x − 3x + 12

= x(x − 4) − 3x + 12 by factorizing the first two terms

= x(x − 4) − 3(x − 4) by factorizing the last two terms extracting a factor of −3 in order to leave x − 4

= (x − 4)(x − 3) by noting the common factor of x − 4 Once again, note that the answer can be checked by multiplying out the brackets again.

Example 2:

Suppose we wish to factories the quadratic expression x2 − 5x − 14. Starting as before we look for two numbers which multiply together to give −14 and add together to give −5. Think about this for a minute and you will realize that the two numbers we seek are −7 and 2 because

−7 × 2 = −14, and − 7+2= −5

So, using the two numbers which add to give −5 we split the −5x term into −7x and +2x. We set the calculation out like this:

x2 − 5x − 14 = x2 − 7x + 2x − 14

= x(x − 7) + 2x − 14 by factorizing the first two terms

= x(x − 7) + 2(x − 7) by factorizing the last two terms

= (x − 7)(x + 2) by noting the common factor of x − 7

So the factorization of x2 − 5x − 14 is (x − 7)(x + 2).

## Factorising by inspection

Clearly, when you have some experience of this method, it is possible to avoid writing out all these stages. This is rather long-winded. What you need to do now is carry out the process by inspection. That is, simply by looking at the given expression, decide in your head which numbers need to be placed in the brackets to do the job. Look at the following two examples and see if you can do this.

Example :

Suppose we want to factorize the quadratic expression x^2 − 9x + 20 by inspection. We write x^2 − 9x + 20 = ( )( )

and try to place the correct quantities in brackets. Clearly we will need an x in both terms:

x^2 − 9x + 20 = (x )(x )

We want two numbers which multiply to give 20 and add to give −9. With practice you will be able to do this in your head. The two numbers are −4 and −5. x^2 − 9x + 20 = (x − 4)(x − 5) The answer should always be checked by multiplying-out the brackets again.

Example:

Suppose we want to factorise the quadratic expression x^2 − 9x − 22 by inspection. We write

x^2 − 9x − 22 = ( )( )

and try to place the correct quantities in brackets. Clearly we will need an x in both terms:

x^2 − 9x − 22 = (x )(x )

We want two numbers which multiply to give −22 and add to give −9. The two numbers are −11 and +2

x^2 − 9x − 22 = (x − 11)(x + 2)

The answer should always be checked by multiplying-out the brackets again. If you can’t manage to do these by inspection yet do not worry. Do it the way we did it before. Your ability to do this will improve with practice and experience. All of these examples have involved quadratic expressions where the coefficient of x^2 was 1. When the coefficient is a number other than 1 the problem is more difficult. We will look at examples like this in the next section.

## Expressions where the coefficient of x^2 is not 1.

Suppose we wish to factorize the expression 3x^2 + 5x − 2. As we did before we look for two numbers which add to give the coefficient of x; so we seek two numbers which add to give 5. However, instead of looking for two numbers which multiply to give −2 we must look for two numbers which multiply to give −6, (that is, the coefficient of x^2 multiplied by the constant term, 3 × −2). This is entirely consistent with the method we applied before because in those examples the coefficient of x^2 was always 1.

So

? × ? = −6

? + ?=5

By inspection, or trial and error, two such numbers are 6 and −1.

6 × −1 = −6

6+ −1=5

We use these two numbers to split the 5x term into 6x and −1x.

3x^2 + 5x − 2

=3x2 + 6x − x − 2

= 3x(x + 2) − x − 2 by factorizing the first two terms

= 3x(x + 2) − (x + 2) factorizing the last two terms by extracting −1

= (x + 2)(3x − 1) by noting the common factor of x + 2

Example 1:

Suppose we wish to factorize 2x^2 + 5x − 7. As we did before we look for two numbers which add to give the coefficient of x; so we seek two numbers which add to give 5. We look for two numbers which multiply to give −14, (that is, the coefficient of x^2 multiplied by the constant term, 2 × −7).

? × ? = −14

? + ? = 5

By inspection, or trial and error, two such numbers are 7 and −2.

7 × −2 = −14
7 + −2=5

We use these two numbers to split the 5x term into 7x and −2x. 2x^2 + 5x − 7

=2x^2 + 7x − 2x − 7

= x(2x + 7) − 2x − 7 by factorising the first two terms

= x(2x + 7) − (2x + 7) factorising the last two terms by extracting −1

= (2x + 7)(x − 1) by noting the common factor of 2x + 7

## Common Factoring

Greatest common factor – largest quantity that is a factor of all the integers or polynomials involved

Finding the GCF of a List of Integers or Terms

1)Prime factor the numbers.

2)Identify common prime factors.

3)Take the product of all common prime factors.

If there are no common prime factors, GCF is 1

## Examples

1)6, 8 and 46

6 = 2 · 3

8 = 2 · 2 · 2

46 = 2 · 23

So the GCF is 2.

2)144, 256 and 300

144 = 2 · 2 · 2 · 3 · 3

256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2

300 = 2 · 2 · 3 · 5 · 5

So the GCF is 2 · 2 = 4.

a3b2, a2b5 and a4b7

a3b2 = a · a · a · b · b

a2b5 = a · a · b · b · b · b · b

a4b7 = a · a · a · a · b · b · b · b · b · b · b

So the GCF is a · a · b · b = a2b2

Notice that the GCF of terms containing variables will use the smallest exponent found amongst the individual terms for each variable.

The first step in factoring a polynomial is to find the GCF of all its terms.

Then we write the polynomial as a product by factoring out the GCF from all the terms.

The remaining factors in each term will form a polynomial.

Factor out the GCF in each of the following polynomials.

6x3 – 9x2 + 12x =

3 · x · 2 · x2 – 3 · x · 3 · x + 3 · x · 4 =

3x(2x2 – 3x + 4)

2) 14x3y + 7x2y – 7xy =

7 · x · y · 2 · x2 + 7 · x · y · x – 7 · x · y · 1 =

7xy(2x2 + x – 1)

Remember that factoring out the GCF from the terms of a polynomial should always be the first step in factoring a polynomial.

This will usually be followed by additional steps in the process.

Factor 90 + 15y2 – 18x – 3xy2.

90 + 15y2 – 18x – 3xy2 = 3(30 + 5y2 – 6xxy2) =

3(5 · 6 + 5 · y2 – 6 · x – x · y2) =

3(5(6 + y2) – x (6 + y2)) =

3(6 + y2)(5x)

Factoring Simple Trinomials

• A simple trinomial is a polynomial with three terms of the form x2+bx+c.

• It is called a simple trinomial because the squared term has a coefficient of 1.

• These trinomials are formed from multiplying two binomials together. eg. Multiply (x+3)(x+4)

• Factoring is the reverse of this process, we will go from x^2+7x+12 to (x+3)(x+4).

• Note that 3 and 4 multiply to 12 and add to 7.

## Complex Trinomials

To work backward and factor a complex trinomial, we must determine the two values, p and q, with a sum of b and a product of ac. To make things work, we decompose the middle term into separate two terms with coefficients p and q. This process is often referred to as decomposition. The resulting expression will look like this: ax2 + bx+ c Since ac, b and c all share the same factors, it will be possible to use grouping to find common factors. Some examples should make this clearer.

Factor 2x ^2 + 13x + 15.

We must determine two numbers, p and q, that have a product of 2 × 15 = 30 and a sum of 13. Since the product and sum are both positive, both p and q will be positive. Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. The two numbers that meet the requirements are 3 and 10.

Decompose the middle term into two terms with p and q as coefficients. 2x^ 2 + 13x + 15 = 2x^ 2 + 3x + 10x + 15 Factor by grouping. 2x^ 2 + 3x + 10x + 15

= x(2x + 3) + 5(2x + 3)

= (2x + 3)(x + 5) Therefore, 2x 2 + 13x + 15

= (2x + 3)(x + 5).

Factor 6x^ 2 + 19x − 7. We must determine two numbers, p and q, that have a product of 6(−7) = −42 and a sum of 19. Since the product is negative and the sum is positive, p will be positive and |p| > |q|. Factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42. The two numbers that meet the requirements are 21 and −2.

Decompose the middle term into two terms with p and q as coefficients. 6x^ 2 + 19x − 7 = 6x 2 + 21x − 2x − 7 Factor by grouping. 6x 2 + 21x − 2x − 7

= 3x(2x + 7) − 1(2x + 7)

= (2x + 7)(3x − 1) Therefore, 6x 2 + 19x − 7

= (2x + 7)(3x − 1).

Factor 25^ 2 − 20x + 4. We must determine two numbers, p and q, that have a product of 25 × 4 = 100 and a sum of −20. Since the product is positive and the sum is negative, both p and q will be negative. Factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50 and 100. In this case, the two numbers that meet the requirements have the same value: −10 and −10.

Decompose the middle term into two terms with p and q as coefficients. 25x^ 2 − 20x + 4 = 25x 2 − 10x − 10x + 4 Factor by grouping. 25x 2 − 10x − 10x + 4

= 5x(5x − 2) − 2(5x − 2)

= (5x − 2)(5x − 2) Therefore, 25x 2 − 20x + 4

= (5x − 2)(5x − 2), or (5x − 2)2 . While decomposition works for perfect squares like the one above, we will look at a more efficient method in the next lesson

## Difference Of Two Squares

A special case concerns what is known as the difference of two squares. What does this look like ? Typically, x^2 − 9. Note that there is no x term and that the number 9 is itself a square number. A square number is one which has resulted from squaring another number. In this case 9 is the result of squaring 3, (3^2 = 9), and so 9 is a square number. Hence x^2 − 9 is the difference of two squares, x^2 − 3^2. When we try to factorize x^2 − 9 we are looking for two numbers which add to zero (because there is no term in x), and which multiply to give −9. Two such numbers are −3 and 3 because

−3 × 3 = −9, and − 3+3=0

We use these two numbers to split the 0x term into −3x and 3x

x^2 − 9

= x^2 − 3x + 3x − 9

= x(x − 3) + 3x − 9 by factorizing the first two terms

= x(x − 3) + 3(x − 3) factorizing the last two terms by extracting +3

= (x − 3)(x + 3) by noting the common factor of x − 3

x^2 − 9=(x − 3)(x + 3) Recall that 9 is a square number, (9 = 3^2). So this example is a difference of two square. The result of this example is true in more general cases of the difference of two squares. It is always the case that x^2 − a^2 factorizes to (x − a)(x + a).

Example:

Suppose we want to factorizes x^2 − 25.

Note that x^2 − 25 is the difference of two squares because 25 is a square number (25 = 52). So we need to factorize x^2 − 52. We can do this by inspection using the formula above. x^2 − 52 = (x − 5)(x + 5)

## complete squares

A second important special case is when we need to deal with complete squares. This happens when the answer can be written in the form ( )^2, that is as a single term, squared.

Suppose we wish to factorize x^2 + 10x + 25. As before we want two numbers which multiply to give 25 and add to give 10.

Two such numbers are 5 and 5 because 5 × 5 = 25, and 5 + 5 = 10 We use these two numbers to split the 10x term into 5x and 5x.

x^2 + 10x + 25

= x^2 + 5x + 5x + 25

= x(x + 5) + 5x + 25 by factorizing the first two terms

= x(x + 5) + 5(x + 5) factorizing the last two terms by extracting +5

= (x + 5)(x + 5) by noting the common factor of x + 5 The result can be written as (x + 5)2, a complete square.

Example:

Suppose we want to factories 25x^2 − 20x + 4. Note that 25x^2 can be written as (5x)^2, a squared term.

Note also that 4 = 22. In this case, by inspection,

25x^2 − 20x + 4 = (5x − 2)(5x − 2)

The result can be written as (5x − 2)^2, a complete square

Do not worry if you have difficulty with this last example. The skill will come with practice. Even if you did not recognize the complete square, you could still use the previous method: To factorize 25x^2 − 20x + 4

As before we want two numbers which multiply to give 100 (the coefficient of x^2 multiplied by the constant term) and add to give −20.

Two such numbers are −10 and −10 because −10 × −10

= 100, and − 10 + −10

= −20 We use these two numbers to split the −20x term into −10x and −10x.

25x2 − 20x + 4

= 25x2 − 10x − 10x + 4

= 5x(5x − 2) − 10x + 4 by factorising the first two terms

= 5x(5x − 2) − 2(5x − 2) factorising the last two terms by extracting −2

= (5x − 2)(5x − 2) by noting the common factor of 5x − 2

So 25x^2 − 20x + 4

= (5x − 2)^2 as before.

## The constant term is missing

There is one more special case that you ought to be familiar with. This arises in examples where the constant term is absent, as in 3x^2 − 8x. How do we proceed ? The way forward is to look for common factors. In the case of 3x^2 −8x note that each of the two terms has a common factor of x. (Note that this is not the case when dealing with a quadratic expression containing a constant term). The common factor, x, is written outside a bracket, and the contents of the bracket are chosen by inspection so that they multiply-out to give the correct terms. 3x^2 − 8x = x(3x − 8)

## Factoring My Grouping

Note that the polynomial is factored by grouping the first and second terms and the third and fourth terms. You could just as easily have grouped the first and third terms and the second and fourth terms, as follows.

## Graphing Factored Form

A quadratic equation is a polynomial equation of degree of 2. The standard form of a quadratic equation is

0 = ax^2 + bx + c

where a, b, and c are all real numbers and a ≠ 0.

If we replace 0 with y , then we get a quadratic Function

y = ax^2 + bx + c

whose graph will be a parabola.

The points where the graph intersects the x-axis will be the solutions to the equation, ax^2 + bx + c = 0. That is, if the polynomial ax^2 + bx + c can be factored to ( xp )( xq ), w e know by the zero product principle that if ( xp )( x q ) = 0, either ( x p ) = 0 or ( x q ) = 0. Then p and q are the solutions to the equation ax^2 + bx + c = 0 and therefore the x-intercepts of the quadratic equation.

Since the x-coordinate of the vertex is exactly the midpoint of the x-intercept, the x-coordinate of the vertex will be .

You can use the x-coordinate of the vertex to find the y-coordinate.

Now you have the vertex and 2 other points on the parabola (namely, the x-intercepts). You can use these three points to sketch the graph

## Example 1:

Graph the function using factoring.

Compare the equation with the standard form,y = ax^2 + bx + c . Since the value of a is positive, the parabola opens up.

Factor the trinomial, . Identify 2 numbers whose sum is –8 and the product is 12. The numbers are –2 and –6. That is,

.

So, by the zero product property, either x – 2 = 0 or x – 6 = 0. Then the roots of the equation are 2 and 6.

Therefore, the x-intercepts of the function are 6 and 2.

The x-coordinate of the vertex is the midpoint of the x -intercepts. So, here the x-coordinate of the vertex will be .

Substitute x = 4 in the equation to find the y-coordinate of the vertex.

That is, the coordinates of the vertex are (4, –4).

Now we have 3 points (4, –4), (2,0) and (6,0) which are on the parabola. Plot the points. Join them by a smooth curve and extend the parabola.

## Standard Form Of A Quadratic function

The quadratic equation is a formula which is used to find the zeroes(x-intercepts). It is equivalent to factoring and I recommend it if the equation is not easily factor-able

## Completing The Square

In this unit we consider how quadratic expressions can be written in an equivalent form using the technique known as completing the square. This technique has applications in a number of areas, but we will see an example of its use in solving a quadratic equation. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that all this becomes second nature. To help you to achieve this, the unit includes a substantial number of such exercises.
Completing The Square ~ Sabrina Singh

## Introduction To Standard

In this unit we look at a process called completing the square. It can be used to write a quadratic expression in an alternative form. Later in the unit we will see how it can be used to solve a quadratic equation.

Some simple equations

Example Consider the quadratic equation x^2 = 9. We can solve this by taking the square root of both sides:

x = 3 or − 3 remembering that when we take the square root there will be two possible answers, one positive and one negative. This is often written in the briefer form x = ±3. This process for solving x^2 = 9 is very straightforward, particularly because: • 9 is a ‘square number’, or ‘complete square’. This means that it is the result of squaring another number, or term, in this case the result of squaring 3 or −3. • x^2 is a complete square - it is the result of squaring x. So simply square-rooting both sides solves the problem.

Example Consider the equation x^2 = 5.

Again, we can solve this by taking the square root of both sides: x = √ 5 or − √ 5 In this example, the right-hand side of x^2 = 5, is not a square number. But we can still solve the equation in the same way. It is usually better to leave your answer in this exact form, rather than use a calculator to give a decimal approximation.

Suppose we wish to solve the equation (x − 7)^2 = 3

Again, we can solve this by taking the square root of both sides. The left-hand side is a complete square because it results from squaring x − 7. x − 7

= √ 3 or − √ 3 By adding 7 to each side we can obtain the values for x: x

= 7 + √ 3 or 7 − √ 3

We could write this in the briefer form x = 7 ± √ 3.

Example

Suppose we wish to solve (x + 3)^2 = 5

Again the left-hand side is a complete square. Taking the square root of both sides: x + 3 = √ 5 or − √ 5 By subtracting 3 from each side we can obtain the values for x: x = −3 + √ 5 or − 3 − √ 5

The basic technique

Now suppose we wanted to try to apply the method used in the three previous examples to x^2 + 6x = 4 In each of the previous examples, the left-hand side was a complete square. This means that in each case it took the form (x + a)^2 or (x − a)^2 . This is not the case now and so we cannot just take the square-root. What we try to do instead is rewrite the expression so that it becomes a complete square - hence the name completing the square. Observe that complete squares such as (x + a)^2 or (x − a)^2 can be expanded as follows:

We will use these expansions to help us to complete the square in the following examples.

Example

Consider the quadratic expression x^2 + 6x − 4

We compare this with the complete square x^2 + 2ax + a^2

Clearly the coefficients of x^2 in both expressions are the same. We would like to match up the term 2ax with the term 6x. To do this note that 2a must be 6, so that a = 3.

Recall that

(x + a)^2

= x^2 + 2ax + a^2

Then with a = 3 (x + 3)^2

= x^2 + 6x + 9 This means that when trying to complete the square for x^2 +6x −4

we can replace the first two terms, x^2 + 6x, by (x + 3)62 − 9.

So x^ 2 + 6x − 4

= (x + 3)^2 − 9 − 4

= (x + 3)^2 − 13

We have now written the expression x^ 2 + 6x − 4 as a complete square plus or minus a constant. We have completed the square. It is important to note that the constant term, 3, in brackets is half the coefficient of x in the original expression.

Example

Suppose we wish to complete the square for the quadratic expression x^2 − 8x + 7.

We want to try to rewrite this so that it takes the form of a complete square plus or minus a constant.

We compare x^ 2 − 8x + 7 with the standard form x ^2 − 2ax + a 2

The coefficients of x 2 are the same. To make the coefficients of x the same we must choose a to be 4.

Recall that (x − a) ^2

= x ^2 − 2ax + a 2 Then with a

= 4 (x − 4)^2 = x 2 − 8x + 16

This means that when trying to complete the square for x^ 2 −8x +7 we can replace the first two terms, x 2 − 8x, by (x − 4)^2 − 16.

So x^ 2 − 8x + 7

= (x − 4)^2 − 16 + 7

= (x − 4)^2 − 9 We have now written the expression x^ 2 − 8x + 7 as a complete square plus or minus a constant. We have completed the square. Again note that the constant term, −4, in brackets is half the coefficient of x in the original expression.

Example

Suppose we wish to complete the square for the quadratic expression x^ 2 + 5x + 3. This means we want to try to rewrite it so that it has the form of a complete square plus or minus a constant. In the examples we have just worked through we have seen how this can be done by comparing with the standard forms (x + a)^ 2 and (x − a) ^2 . We would like to be able to ‘complete the square’ without writing down all the working we did in the previous examples. The key point to remember is that the number in the bracket of the complete square is half the coefficient of x in the quadratic expression.

We have now written the expression x^ 2 + 5x + 3 as a complete square plus or minus a constant. We have completed the square. Again note that the constant term, 5/2 , in brackets is half the coefficient of x in the original expression. The explanation given above is really just an outline of our thought process; when we complete the square in practice we would not write it all down. We would probably go straight to equation (1). The ability to do this will come with practice.

Cases in which the coefficient of x 2 is not 1

We now know how to complete the square for quadratic expressions for which the coefficient of x ^2 is 1. When faced with a quadratic expression where the coefficient of x^ 2 is not 1 we can still use this technique but we put in an extra step first - we factor out this coefficient

Suppose we wish to complete the square for the expression 3x ^2 − 9x + 50. We begin by factoring out the coefficient of x ^2 , in this case 3. It does not matter that 3 is not a factor of 50; we can still do this by writing the expression as

and we have completed the square. This is the ‘completing the square’ form for a quadratic expression for which the coefficient of x ^2 is not 1.

Summary of the process

Solving a quadratic equation by completing the square

Let us return now to a problem posed earlier. We want to solve the equation x^ 2 + 6x = 4. We write this as x ^2 + 6x − 4 = 0. Note that the coefficient of x^ 2 is 1 so there is no need to take out any common factor. Completing the square for quadratic expression on the left-hand side:

We have solved the quadratic equation by completing the square. To produce equation (1) we have noted that the coefficient of x in the quadratic expression is 6 so the number in the ‘complete square’ bracket must be 3; then we have balanced the constant by subtracting the square of this number, 3 ^2 , and putting in the constant from the quadratic, −4. To get equation (2) we just do the arithmetic which in this example is quite straightforward.

We are now ready to derive and use the quadratic formula, which will allow us to solve all quadratic equations. We derive the formula by using the method of completing the square. To use the quadratic formula, the quadratic equation you want to solve must be in standard form. That form is ax^2+ bx+ c= 0 in which a is not 0
This unit is about the solution of quadratic equations. These take the form ax^2 +bx+c = 0. We will look at four methods: solution by factorization, solution by completing the square, solution using a formula, and solution using graphs In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial on this topic, you should be able to:

• solve quadratic equations by factorisation

• solve quadratic equations by completing the square

• solve quadratic equations using a formula

• solve quadratic equations by drawing graphs

This unit is about how to solve quadratic equations. A quadratic equation is one which must contain a term involving x^ 2 , e.g. 3x ^2 , −5x^ 2 or just x ^2 on its own. It may also contain terms involving x, e.g. 5x or −7x, or 0.5x. It can also have constant terms - these are just numbers: 6, −7, 1/ 2 . It cannot have terms involving higher powers of x, like x^ 3 . It cannot have terms like 1 x in it. In general a quadratic equation will take the form ax^2 + bx + c = 0 a can be any number excluding zero. b and c can be any numbers including zero. If b or c is zero then these terms will not appear.

In this section we will assume that you already know how to factorize a quadratic expression. If this is not the case you can study other material in this series where factorization is explained.

Example

Suppose we wish to solve 3^x 2 = 27. We begin by writing this in the standard form of a quadratic equation by subtracting 27 from each side to give 3x^ 2 − 27 = 0.

We now look for common factors. By observation there is a common factor of 3 in both terms. This factor is extracted and written outside a pair of brackets. The contents of the brackets are adjusted accordingly: 3x^ 2 − 27

= 3(x^ 2 − 9) = 0 Notice here the difference of two squares which can be factorized as 3(x ^2 − 9) = 3(x − 3)(x + 3) = 0

If two quantities are multiplied together and the result is zero then either or both of the quantities must be zero. So either x − 3 = 0 or x + 3 = 0

so that x = 3 or x = −3

These are the two solutions of the equation.

Solving quadratic equations by completing the square

Again we have two answers. These are exact answers. Approximate values can be obtained using a calculator.
Solving quadratic equations using a formula

Consider the general quadratic equation ax^2 + bx + c = 0.

There is a formula for solving this: x = −b ± √ b^ 2 − 4ac /2a . It is so important that you should learn it.

We will illustrate the use of this formula in the following example.

Example

Suppose we wish to solve x^ 2 − 3x − 2 = 0. Comparing this with the general form ax^2 + bx + c = 0 we see that a = 1, b = −3 and c = −2. These values are substituted into the formula

These solutions are exact

Suppose we wish to solve 3x ^2 = 5x − 1. First we write this in the standard form as 3x^ 2 − 5x + 1 = 0 in order to identify the values of a, b and c. We see that a = 3, b = −5 and c = 1. These values are substituted into the formula.

Again there are two exact solutions. Approximate values could be obtained using a calculator

Solving quadratic equations by using graphs

In this section we will see how graphs can be used to solve quadratic equations. If the coefficient of x 2 in the quadratic expression ax^2 + bx + c is positive then a graph of y = ax^2 + bx + c will take the form shown. If the coefficient of x ^2 is negative the graph will take the form .

Graphs of y = ax^2 + bx + c have these general shapes We will now add x and y axes. Figure 2 shows what can happen when we plot a graph of y = ax^2 + bx + c for the case in which a is positive.
Graphs of y = ax^2 + bx + c when a is positive

The horizontal line, the x axis, corresponds to points on the graph where y = 0. So points where the graph touches or crosses this axis correspond to solutions of ax^2 + bx + c = 0. In Figure 2, the graph in (a) never cuts or touches the horizontal axis and so this corresponds to a quadratic equation ax^2 + bx + c = 0 having no real roots. The graph in (b) just touches the horizontal axis corresponding to the case in which the quadratic equation has two equal roots, also called ‘repeated roots’. The graph in (c) cuts the horizontal axis twice, corresponding to the case in which the quadratic equation has two different roots

## Graphing Standard Form

STANDARD FORM: y = ax^2 + bx + c

Vertex: find the x-value of the vertex by using the handy dandy formula –b/2a Once you find the x-value, plug it into the equation to find the y-value.

Axis of symmetry: This is the line that is smack in the middle of the parabola – it is the mirror for a parabola. In standard form it is a vertical line in the form: x = –b/2a

y-intercept: This is an easy one – like always we find the y-intercept by plugging in x = 0. In this case that makes the y-intercept (0, c)

Other points: Plug in an x-value on one side or another of the vertex into the equation. If you know where the axis of symmetry is then you get some points for free by finding reflections over the axis of symmetry. Plot a point, count horizontally to the axis of symmetry, then keep on going the same number and place a point.

Ex: y = x^2 – 4x - 5

a = 1, b = -4, c = 5

Vertex x = 4 /2(1) = 2 plug in x = 2 Y = (2)2 -4(2) - 5 = 4 - 8 – 5 = -9 (2, -9)

axis of symmetry is x = 2 (through the vertex)

y – intercept (0, -5)

Plug in another point like x = -1 Which gives us (-1, 0)

Now we can count from the axis of symmetry From our two points to get two additional points: From (0,-5) to the axis you must travel 2 spaces to the right, so keep going two more spaces to get the point (4, -5). From (-1, 0) you must travel 3 spaces to the right, continue on 3 more spaces and come to (5, 0). Now sketch the graph through those points.

## Motion Problem

The height, , in feet of an object above the ground is given by h=-16t^2+64+190 where t is the time in seconds. Find the time it takes the object to strike the ground and find the maximum height of the object.

Let’s first find the time it takes for the object to hit the ground. Since represents the height above the ground, we would like to know at what time h=0. So in the equation h=-16t^2+64+190 we will set h=0 and solve for t.

h=-16t^2+64+190

0=-16t^2+64+190

So we simply want to solve this quadratic equation. It is easiest to use the quadratic formula in this situation. So we get

Therefore you get 5.98, -1.98

However, since t represents time, we must throw out –1.98. Therefore, it takes 5.98 seconds for the object to strike the ground.

The other part of the question is we want to know that maximum height that the object reaches. Since we can see that the function is clearly a quadratic function which opens down, we know that this maximum must occur at the vertex. So let’s find the vertex

t= -b/2a

so we get - 64/2(16)

=-16/-32

=2

So at t=2 seconds the object reaches its maximum height. However, we wanted to know what that maximum height is. Therefore, we must find the value of the vertex, in this case it will be the value h of t=2 when . So we plug this in to get

h=-16(2)^2+64(2)+190

= 254

Therefore the maximum is 254 feet.

## Revenue Problem

Revenue = (number of items sold) x (price of per item)

Castlebrooke's Athletic Council is selling tickets for their first ever FNL Football game.The current price of an amateur theater tickets is \$20, and the venue typically sells 500 tickets. A survey found that for each \$1 increase in ticket price, 10 fewer tickets are sold.

(a) What is the number of \$1 increases in price that will maximize the revenue?

(b) What price per ticket will maximize the revenue?

(c) What is the maximum revenue?

These parts asks for the number of \$1 increases, so, let x be the number of \$1 increases. The current revenue before the survey is R = (500) (20) because there are 500 items sold at \$20 each. Now add the survey results into the equation using x as the number of \$1 increases

• The price increases for each \$1 increments, so the price changes from (20) to (20 + 1x)
• For each \$1 increase, 10 fewer tickets are sold. Tickets sold changes from (500) to (500 - 10x).

The revenue equation based on the survey is R = (500 - 10x) (20 + 1x). Because the question says “maximize” we need to find the vertex of R = (500 - 10x) (20 + 1x) . Since the equation is already factored (although not fully factored) the fastest way to do this is by using the midpoint of the zeros. Solve each bracket for zero just like any other quadratic equation. The first zero is 500÷10=50, and the second is -20. The midpoint of these zeros is (50-20)÷2=15. This means that h = 15, and the x-value of the vertex is 15. In short, x = 15.

(a): The theatre should have 15, \$1 increases to maximize their revenue.

(b): The new price will be 20 + 1×15, or \$35 per ticket.

For part (c) we need to find the k value of the vertex. To do this, substitute x = 15 in the R equation, and calculate R. R=(500-10×15) (20+1×15)=12250

(c): maximum revenue will be \$12250.

## Numeric Problems

The sum of the squares of two consecutive integers is 365. What are the integers?

Let one number be x. Since the two numbers are consecutive, the other number is x + 1.

The sum of the squares is (x)^2 + (x + 1)^ 2 and it is equal to 365. The equation is (x) 2 + (x + 1)^2 = 365.

If you look at the question, it does not say “maximum”, “minimum” so we do not need the vertex. What we need to do, is to solve the equation (x)^2 + (x + 1)^2 = 365, by factoring or quadratic formula.

Expand and simplify the equation into standard form: 2x^2 + 2x – 364 = 0 Factor or use the quadratic formula to find the two solutions: x = -14, x = 13.

The two solutions to the equation will give us two possible sets of integers which are answers to the question: The first set of integers is -14 and -13, and the second set of integers is 13 and 14.

## Reflection

I don't really have any of my test of quizzes from quadratics but that doesn't mean I can't tell you what I've learned throughout the Quadratics from where I started to where I ended off. When coming into quadratics I had never even heard of the word. Now I know about the different degrees of functions. Which is when a polynomial is expressed as a sum or difference of terms, the term with the highest degree is the degree of the polynomial. The degree of a term is the sum of the powers of each variable in the term. In functions of a single variable, the degree of a term is simply the exponent. From learning about degree of difference of functions to the difference between successive terms of a sequence form another sequence of numbers. If these numbers are constant, the sequence is a 1st level difference. If these numbers are not constant and their difference form a sequence of constant numbers, then the sequence is a 2nd level difference. After we passed this and we went onto graphing I was able to understand how to do so because each for had a specific format which you had to follow and if you did that you will graph it right all the time. This is something I liked because finding the x-intercepts, axis of symmetry, vertex, and optimal value was very easy in my opinion.

## Math Learning Goals ( mini reflection of each part of quadratics)

Using Algebra tiles

• Connect the algebraic representations to the graphical representations of quadratics of the forms y = x 2 + bx + c and y = (x – r) (x – s).
• Expand and simplify second-degree polynomial expressions involving one variable that consist of the product of two binomials or the square of a binomial, using algebra tiles.

Multiply a Binomial

• Expand and simplify second-degree polynomial expressions involving one variable that consist of the product of two binomials or the square of a binomial, using the chart method, and the distributive property.

Find the y-Intercept of Quadratic Equations

• Determine the connections between the factored form of a quadratic relation and the x-intercepts of its graph.
• Factor simple trinomials of the form x 2 + bx + c, using a variety of tools, e.g., algebra tiles, paper and pencil, and patterning strategies.