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4Quadratic Functions

Copyright © Cengage Learning. All rights reserved.

4.4 Solving Quadratic Equations by the Square RootProperty and Completing the Square

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Objectives

Solve a quadratic equation using the square root property.

Solve a quadratic equation by completing the square.

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Solving from Vertex Form

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Solving from Vertex Form

One characteristic of numbers to be cautious about is that when we square a real number, squaring will effectivelyremove any negative sign.

x2 = 25

(5)2 = 25 (–5)2 = 25

x = 5 x = –5

x2 = 25

x =

x = 5 x = –5

We know that 5 squared is 25.Negative 5 squared is also 25.

Both must be given as answers to this equation.

When using a square root, we must use the plus/minus symbol to represent both answers.

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Solving from Vertex Form

When we write a square root, we are referring to the positive solution, which is called the principal square root.Therefore, = 5.

Because the principal square root only accounts for positive solutions we lose possible negative results, so we must use a plus/minus symbol () to show that there are two possible answers.

Using the square root and plus/minus is an example of the square root property.

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Solving from Vertex Form

With this property in mind, we are going to look at two basicways to solve quadratic equations: the square root propertyand completing the square. First, we will use the squareroot property to solve quadratics when they are given invertex form.

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Solving from Vertex Form

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Example 1 – Solving quadratic equations using the square root property

Solve 10 = 2(x – 4)2 – 8.

Solution:

10 = 2(x – 4)2 – 8

+8 +8

18 = 2(x – 4)2

9 = (x – 4)2

Isolate the squared variable expression.

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Example 1 – Solution

3 = x – 4

3 = x – 4 or –3 = x – 4

7 = x or 1 = x

10 ≟ 2(7 – 4)2 – 8

10 = 10

Rewrite as two equations andsolve.

Check both answers.

Use the square root property. Don’t forget the plus/minus symbol.

cont’d

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Example 1 – Solution

10 ≟ 2(1 – 4)2 – 8

10 = 10

x = 7 and x = 1 are both valid answers to this equation.

Both answers work.

cont’d

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Completing the Square

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Completing the Square

If a quadratic equation has both a second-degree term anda first-degree term (ax2 + bx + c), the square root property cannot be easily used to solve.

If we try to get the squared variable expression by itself, we will have a variable term in the way of the square root. To handle this problem, we use a technique called completing the square.

Completing the square will transform the equation so that ithas a perfect square that can be solved using the squareroot property.

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Completing the Square

In general, taking half of the coefficient of the first-degree term and squaring it will give us a constant that will makethese expressions a perfect square trinomial.

x2 + bx

The factored form will include half of b as one of the terms in the binomial.

The constant that will complete

the square:

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Completing the Square

We will start the process of completing the square by simplifying one side of an equation.

This process is easiest when the coefficient of the squared term is 1. If this coefficient is not 1, we will divide both sides of the equation by the coefficient to make it 1.

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Example 4 – Solving quadratic equations by completing the square

Solve by completing the square of 5x2 + 30x – 35 = 0.

Solution:

Step 1: Isolate the variable terms on one side of the equation.

To isolate the variable terms, we move the constant term to the other side of the equation.

5x2 + 30x – 35 = 0

35 35

5x2 + 30x = 35

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Example 4 – Solution cont’d

Step 2: If the coefficient of x2 is not 1, divide both sides of the equation by the coefficient of x2.

Divide both sides by the coefficient, 5.

x2 + 6x = 7

Step 3: Take half of the coefficient of x, then square it. Add this number to both sides of the equation.

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Example 4 – Solution cont’d

To complete the square on the left side of the equation, we add a number that will make the resulting trinomial a perfect square trinomial.

To find this number, we take half of the coefficient of x and square it. This number is then added to both sides of the equation. In this equation, the coefficient of x is 6, so we get

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Example 4 – Solution cont’d

Add this constant to both sides of the equation.

x2 + 6x + 9 = 7 + 9

x2 + 6x + 9 = 16

Step 4: Factor the quadratic into the square of a binomial. The left side will factor as a perfect square.

x2 + 6x + 9 = 16

(x + 3)2 = 16

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Step 5: Solve using the square root property.

(x + 3)2 = 16

x + 3 = 4

x + 3 = 4 x + 3 = –4

x = 1 x = –7

Example 4 – Solution

Use the square root property.

Rewrite as two equationsand solve.

cont’d

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Step 6: Check your answers in the original equation.

5(1)2 + 30(1) – 35 ≟ 0

(5 + 30 – 35) ≟ 0

0 = 0

Example 4 – Solution cont’d

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Example 4 – Solution

(5(–7)2 + 30(–7) – 35) ≟ 0

(5(49) – 210 – 35) ≟ 0

(245 – 210 – 35) ≟ 0

0 = 0

Both answers work.

cont’d

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Completing the Square

Completing the square will allow us to solve any quadraticequation using the square root property.

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Converting to Vertex Form

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Converting to Vertex Form

To put a quadratic function into vertex form, use thetechnique of completing the square.

The process is very similar, but instead of adding the constant that will complete the square to both sides, you will add and subtract the constant on one side.

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Example 6 – Converting a quadratic function into vertex form

Convert to vertex form.

a. f (x) = 2x2 + 20x – 6 b. g(x) = 5x2 + 3x + 20

Solution:

a. Converting a function to vertex form will be similar to completing the square, as we did earlier. Because we are not solving, we will keep everything on one side of the equation, leaving the function notation alone.

f (x) = 2x2 + 20x – 6

f (x) = (2x2 + 20x) – 6

Group the variable terms.

Factor out the coefficientof the squared term.

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Example 6 – Solution cont’d

f (x) = 2(x2 + 10x) – 6

Now we want to complete the square of the variable term

in parentheses. Find the constant that will complete the square.

52 = 25

Add and subtract 25 inside the parentheses. Then take out the subtracted 25, keeping the multiplication by 2. Finally, you can simplify and factor.

f (x) = 2(x2 + 10x + 25 – 25) – 6

Add and subtract 25 inside the parentheses.

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Example 6 – Solution

f (x) = 2(x2 + 10x + 25) – 2(25) – 6

f (x) = 2(x2 + 10x + 25) – 50 – 6

f (x) = 2(x + 5)2 – 56

b. g(x) = 5x2 + 3x + 20

g(x) = (5x2 + 3x) + 20

Bring out the subtracted 25, keeping the multiply by 2 from the parentheses.Simplify and factor.

Group the variable terms.

cont’d

Factor out the coefficient of the squared term.

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Find the constant that will complete the square.

Example 6 – Solution

Add and subtract

inside the parentheses.

cont’d

Bring out the subtracted , keeping the multiply by 5 from the parentheses.

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Example 6 – Solution

Simplify.

Factor.

cont’d

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Graphing from Vertex Form with x-Intercepts

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Graphing from Vertex Form with x-Intercepts

Now that we know how to solve a quadratic in vertex form,we can find the horizontal intercepts of a quadratic in vertexform, and these can be part of our graphs.

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Example 7 – Graphing a quadratic function in vertex form including horizontal intercepts

Sketch the graph of f (x) = –1.5(x – 2.5)2 + 45.375. Give thedomain and range of the function.

Solution:

Step 1: Determine whether the graph opens up or down.

The value of a is –1.5. Because a is negative, the graph will open downward.

Step 2: Find the vertex and the equation for the axis of symmetry.

Because the quadratic is given in vertex form, the vertex is (2.5, 45.375).The axis of symmetry is the vertical line through the vertex x = 2.5.

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Example 7 – Solution cont’d

Step 3: Find the vertical intercept.

To find the vertical intercept, we make the input variable zero.

f (x) = –1.5(x – 2.5)2 + 45.375

f (0) = –1.5(0 – 2.5)2 + 45.375

f (0) = –1.5(–2.5)2 + 45.375

f (0) = –1.5(6.25) + 45.375

f (0) = 36

Therefore, the vertical intercept is (0, 36).

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Example 7 – Solution

Step 4: Find the horizontal intercepts (if any).

Horizontal intercepts occur when the output variable is equal to zero, so substitute zero for the output variable and solve.

30.25 = (x – 2.5)2

cont’d

Isolate the squared variable expression.

Use the square root property.

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Example 7 – Solution

5.5 = x – 2.5

5.5 = x – 2.5 –5.5 = x – 2.5

8 = x –3 = x

Therefore, we have the horizontal intercepts (–3, 0) and (8, 0).

cont’d

Write two equations and solve.

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Example 7 – Solution

Step 5: Plot the points you found in steps 2 through 4. Plottheir symmetric points and sketch the graph. (Find an additional pair of symmetric points if

needed.)

The vertical intercept (0, 36)is 2.5 units to the left of the axis of symmetry x = 2.5, so its symmetric point will be 2.5 units to the right of the axis of symmetry at (5, 36).

cont’d

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Example 7 – Solution

Connect the points with a smooth curve. The input variable x can be any real number without causing the function to be undefined, so the domain of thefunction is all real numbers or ( , ).

The graph decreases tonegative infinity, and thehighest output, y = 45.375, is at the vertex so the rangeis ( , 45.375] or y 45.375.

cont’d