Unit 4 Topic 3: SolvingQuadratic Equations

Algebra 1 Summit

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AUTHORAlgebra 1 Summit

www.ck12.org Chapter 1. Unit 4 Topic 3: Solving Quadratic Equations

CHAPTER 1 Unit 4 Topic 3: SolvingQuadratic Equations

Solving Quadratic Equations

There are multiple ways to solve a quadratic equation. Not every method works for every equation, so it is importantto be fluent in multiple methods. Some methods will be easier than others depending on the format of the equation.

The solutions to a quadratic equation are when the function’s y values are equal to zero. A quadratic function canhave one, two or no real solutions.

Method 1: The Table Method

This method is as simple as looking at the table of values for the solutions. Input the equation into the graphingcalculator in y= then select 2nd Graph to find the table of values. The solutions will be the x values which correspondto the y values of zero.

Example 1: x2 + x−6 = 0

TABLE 1.1:

x y-4 6-3 0-2 -4-1 -60 -61 -42 03 6

The solutions of this equation are x=-3 and x=2, because they are the x values, which correspond with the y value ofzero.

Although this method seems very simple, it does not work for every example.

Example 2: x2 +3x−5 = 0

This is the table of values for this function.

TABLE 1.2:

x y-5 5-4 -1-3 -5-2 -7-1 -70 -51 -1

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Method 2: The Graphing Method

The solutions of this equation are not visible in this table and therefore this method cannot be used. It can beestimated that one solution is between -4 and -5 and the other is between 1 and 2, but another method must be usedto find the exact solutions.

This method is as simple as looking at the graph of the function to find the solutions. The solutions will be the xvalues which correspond to the y values of zero. Look to the x axis for the solutions.

FIGURE 1.1

http://jodzilla.wikispaces.com/.

The solutions of this equation are x = -1 and x = 3 because these are the x values which correspond with the y valueof zero. The solutions are also called the “zeros” for this reason.

Example 2: −x2 +2x+2 = 0

Although this method seems very simple, it does not work for every example without a calculator.

FIGURE 1.2

http://www.mathamazement.com/Lessons/Pre-Calculus/02_Polynomial-and-Rational-Functions/quadratic-functions.html

The exact solutions of this equation are not visible on this graph and therefore this method cannot be used. Itcan be estimated that one solution is between -1 and 0 and the other solution is between 2 and 3, but anothermethod must be used to find the exact solutions.

In order to solve this using a graph, use of the graphing calculator is required.

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www.ck12.org Chapter 1. Unit 4 Topic 3: Solving Quadratic Equations

Move curser near these locations.

To find the solutions, hit 2nd Trace.Select 5: intersectMove the curser along the curve towards one of the zeros. When the curser is near the height y = 0, hit enter 3 times.Repeat these steps for the other side.The solutions are x = -0.732 and x = 2.732

Method 3: The Zero Product Property

The Zero Product Property states that for any factors a · b = 0, then either a = 0, b = 0, or they are both zero.

Video on Factoring and ZPP:

https://www.khanacademy.org/math/algebra/quadratics/factoring_quadratics/v/Example%201:%20Solving%20a%20quadratic%20equation%20by%20factoring

If a quadratic equation can be written in factored form, this method can be used. Since not all quadratic functionscan be factored, this method will not work on every problem.

Example 1: (x + 3)(x – 5) = 0

Since this example includes two factors that are multiplied to make zero, take each factor and set them equal to zeroto solve.

x + 3 = 0 and x – 5 = 0

x = -3 and x = 5

The solutions are -3 and 5.

To check the solutions, try each solution in the original equation.

(-3 + 3)(-3 – 5) = 0 and (5 + 3)(5 – 5) = 0

The values of -3 and 5 both satisfy the equation.

Example 2: x2−4x−21 = 0

In this example, the problem must first be factored to be in the form a · b = 0.

(x + 3)(x – 7) = 0

Now that this example includes two factors that are multiplied to make zero, take each factor and set them equal tozero to solve.

TABLE 1.3:

x + 3 = 0 x - 7 = 0x = -3 x = 7

The solutions are -3 and 7.

To check the solutions, try each solution in the original equation.

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TABLE 1.4:

(-3)2 – 4(-3) – 21 = 0 (7)2 – 4(7) – 21 = 00=0 0=0

The values of -3 and 7 both satisfy the equation.

Example 3: x2 +2x+2 = 0

In this example, the quadratic must first be factored to be in the form a · b = 0. Since this quadratic function cannotbe factored, this method cannot be used.

Method 4: The Square Root Method

The Square Root Method is most often used when a quadratic function is written in vertex form a(x – h)2+ k = 0.The equation must be manipulated into vertex form by completing the square if it is not given in this form.

The first step is to have the perfect square alone on one side of the equation. Eliminate the “k” value by adding25 to the other side.

(x – 2)2 – 25 = 0

Next, take the square root of both sides to cancel the square with the square root. This leaves only (x – 2) onthe left.

(x – 2)2 = 25

There are two numbers that when squared result in 25. Set the quantity (x-2) equal to these two values, 5 and-5. Solve each equation for x to find the two solutions.

TABLE 1.5:

(x – 2) = 5 (x – 2) = -5X -2 +2 = 5+ 2 x - 2 + 2 = -5 + 2x = 7 x = - 3

Solve each equation for x to find the two solutions.

x = 7 and x = -3

Example 2: −2(x−3)2 +12 = 0

The first step is to have the perfect square alone on one side of the equation. Eliminate the “k” value by subtracting12 to the other side.

-2(x – 3)2 +12 = 0

-2(x – 3)2 = -12

Divide each side by -2 to isolate the quantity (x – 2)2 on the left.√

(x – 3)2 =√

6

Next, take the square root of both sides to cancel the square. This leaves only (x -3) on the left and

(x-3) =√

6

There are two numbers that when squared results in 6. Since the number is irrational, leave the value as√

6

Check the solutions by evaluating both values in the original equation. Both solutions should result as a true

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www.ck12.org Chapter 1. Unit 4 Topic 3: Solving Quadratic Equations

FIGURE 1.3

statement. Since these values were rounded, the check may be off by a small margin.

Example 3: x2−2x−5 = 0

This example is not given in vertex form. In order to use this method, the equation must be re-written in vertex formby completing the square. For a review on completing the square, watch the video below.

Completing the Square Video:

https://www.khanacademy.org/math/algebra/quadratics/factoring_quadratics/v/Example%201:%20Solving%20a%20quadratic%20equation%20by%20factoring

FIGURE 1.4

Check the solutions by evaluating both values in the original equation. Both solutions should result as a truestatement. Since these values were rounded, the check may be off by a small margin.

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Method 5: The Quadratic Formula

The Quadratic Formula can be used to solve quadratic functions for their solutions. The formula works because itis simply the formula of a quadratic function in standard form (ax2 + bx + c = 0) solved for x. Students do not haveto derive the formula in Algebra 1, but will need to know how it is used.

The Quadratic Formula: −b±√(b2)−4ac

2a

Step 1: Make sure the problem is solved to the form ax2 + bx + c = 0.

Step 2: Identify the values of a, b and c.

Step 3: Plug in the values of a, b and c into the formula.

Step 4: Simplify using order of operations and your calculator.

Example 1:x2−2x−15 = 0

The solutions are 5 and -3.

Check the solutions in the original equation.

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www.ck12.org Chapter 1. Unit 4 Topic 3: Solving Quadratic Equations

Example 2: 2x2 +6x =−4

FIGURE 1.5

Check the solutions in the original equation.

2(-1)2 + 6(-1) = -4 and 2(-2)2 + 6(-2) = -4

-4 = -4 and -4 = -4

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Example 3: x2– 2x + 15 = 0

Hint!

The value under the square root symbol in the formula is called the discriminant. This number identifies how manysolutions the equation will have. Use the chart below for reference:

TABLE 1.6:

Discriminant Value Number of SolutionsPositive Number Two SolutionsZero One SolutionNegative Number Zero Real Solutions

Examples 1 and 2 had a positive number discriminant value and resulted in two solutions. Example 3 had a negativenumber discriminant value and resulted in zero real solutions. Example 4 below is an example of a discriminantvalue of zero which will result in only one solution value.

Example 4: x2 +16 = 8x

Step 1: This problem must be solved into standard form. Subtract the 8x to the left side and place it as the “b” value.

x2 – 8x + 16 = 0

Step 2: a = 1, b = -8, c = 16The solution is 4.

Check the solution in the original equation.

(4)2 – 8(4) + 16 = 0

0 = 0

Quadratic Formula Video:

https://www.khanacademy.org/math/algebra/quadratics/quadratic-formula/v/quadratic-formula-1

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www.ck12.org Chapter 1. Unit 4 Topic 3: Solving Quadratic Equations

The next two examples will show how to solve an equation using all of the methods discussed in this chapter.

FIGURE 1.6

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FIGURE 1.7

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