Guide To Quadratics
Compiled by: Sabrina Singh
Why the name "Quadratics"?
Answer: These equations are intimately connected with problems about squares and quadrangles. (In fact, the word quadratic is derived from the Latin word quadratus for square.) Questions about quadrangles often lead to quadratic equations. For example, consider the problem: A quadrangle has one side four units longer than the other. Its area is 60 square units. What are the dimensions of the quadrangle?If we denote the length of one side of the quadrangle as x units, then the other must be x + 4 units in length. We must solve the equation: x(x+4)+60 , which is equivalent to solving the quadratic equation x^2 +4x-60+0 Solving quadratic equations, even if not derived from a quadrangle problem, still involves the geometry of four-sided shapes. As we shall see, all such equations can be solved by a process of “completing the square.”
Table Of Contents
- The Algebra Of Quadratics
- Quadratic Function
- Analyzing Quadratics (Second Differences)
- Properties Of A Parabola
Types of Equations
- Factored Form a(x-r)(x-s)
- Vertex Form a(x-h)²+k
- Standard Form ax²+bx+c
- Finding an equation when only given the vertex
- Graphing vertex form
- Factors and zeroes
- Polynomials (factored form to standard form)
- Common factoring
- Factoring simple trinomials
- Factoring complex trinomials
- Difference of squares
- Perfect square
- Factoring by grouping
- Graphing factored form
- Completing the square (standard to vertex)
- Solving using quadratic equation
- Graphing from standard form
Algebra Of Quadratics
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.
Analyzing Quadratic (Second Differences)
Part Of A Parabola
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.
- 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
Scaling the Quadratic
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).
A similar thing happens when you replace x with x − 1, only this time the graph is shifted one unit to the right.
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
Types Of Quadratic Equations
- Vertex form: a(x-h)²+k
- Factored form : a(x-r)(x-s)
- Standard form: ax²+bx+c
Vertex Form of Quadratic Functions
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.
- 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:
Graphing Quadratic Function In Vertex Form
- The graph f(x)= a(x-h)^2+k is the parabola y=x^2 translated an h united horizontally and k units vertically.
- Start with the function in vertex form:
- 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)
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)
Factored Form Of A Quadratic Function
Finding polynomial functions ( factors and zeros)
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.
- 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
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.
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!
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!
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.
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
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.
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.
? × ? = −6
? + ?=5
By inspection, or trial and error, two such numbers are 6 and −1.
6 × −1 = −6
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
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
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
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 – 6x – xy2) =
3(5 · 6 + 5 · y2 – 6 · x – x · y2) =
3(5(6 + y2) – x (6 + y2)) =
3(6 + y2)(5 – x)
• 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.
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
−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).
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)
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.
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
Factoring My Grouping
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 ( x – p )( x – q ), w e know by the zero product principle that if ( x– p )( 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
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
Completing The Square
Introduction To Standard
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.
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:
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.
(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.
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.
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
Summary of the process
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:
The Quadratic Formula
• 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.
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
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.
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
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.
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 .
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
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)
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.
So we simply want to solve this quadratic equation. It is easiest to use the quadratic formula in this situation. So we get
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
so we get - 64/2(16)
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
Therefore the maximum is 254 feet.
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.
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.
Math Learning Goals ( mini reflection of each part of quadratics)
- 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.