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Solve x 2 6 x + 5 = 0 by using the Quadratic Formula.

Solution

.
This equation is in standard form. .
Identify the a, b, c values. .
Write the Quadratic Formula. .
Then substitute in the values of a, b, c. .
Simplify. .
.
.
Rewrite to show two solutions. .
Simplify. .
.
Check.
.
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Solve a 2 2 a 15 = 0 by using the Quadratic Formula.

a = −3 , a = 5

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Solve b 2 + 10 b + 24 = 0 by using the Quadratic Formula.

b = −6 , b = −4

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When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula. If we get a radical as a solution, the final answer must have the radical in its simplified form.

Solve 4 y 2 5 y 3 = 0 by using the Quadratic Formula.

Solution

We can use the Quadratic Formula to solve for the variable in a quadratic equation, whether or not it is named ‘ x ’.

.
This equation is in standard form. .
Identify the a, b, c values. .
Write the Quadratic Formula. .
Then substitute in the values of a, b, c. .
Simplify. .
.
Rewrite to show two solutions. .
Check. We leave the check to you.
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Solve 2 p 2 + 8 p + 5 = 0 by using the Quadratic Formula.

p = −4 ± 6 2

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Solve 5 q 2 11 q + 3 = 0 by using the Quadratic Formula.

q = 11 ± 61 10

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Solve 2 x 2 + 10 x + 11 = 0 by using the Quadratic Formula.

Solution

.
This equation is in standard form. .
Identify the a, b, c values. .
Write the Quadratic Formula. .
Then substitute in the values of a, b, c. .
Simplify. .
.
Simplify the radical. .
Factor out the common factor in the numerator. .
Remove the common factors. .
Rewrite to show two solutions. .
Check. We leave the check to you.
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Solve 3 m 2 + 12 m + 7 = 0 by using the Quadratic Formula.

m = −6 ± 15 3

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Solve 5 n 2 + 4 n 4 = 0 by using the Quadratic Formula.

n = −2 ± 2 6 5

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We cannot take the square root of a negative number. So, when we substitute a , b , and c into the Quadratic Formula, if the quantity inside the radical is negative, the quadratic equation has no real solution. We will see this in the next example.

Solve 3 p 2 + 2 p + 9 = 0 by using the Quadratic Formula.

Solution

This equation is in standard form. .
Identify the a, b, c values. .
Write the Quadratic Formula. .
Then substitute in the values of a, b, c. .
Simplify. .
Simplify the radical. .
We cannot take the square root of a negative number. There is no real solution.
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Solve 4 a 2 3 a + 8 = 0 by using the Quadratic Formula.

no real solution

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Solve 5 b 2 + 2 b + 4 = 0 by using the Quadratic Formula.

no real solution

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The quadratic equations we have solved so far in this section were all written in standard form, a x 2 + b x + c = 0 . Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Solve x ( x + 6 ) + 4 = 0 by using the Quadratic Formula.

Solution

.
Distribute to get the equation in standard form. .
This equation is now in standard form. .
Identify the a, b, c values. .
Write the Quadratic Formula. .
Then substitute in the values of a, b, c. .
Simplify. .
Simplify inside the radical. .
Simplify the radical. .
Factor out the common factor in the numerator. .
Remove the common factors. .
Rewrite to show two solutions. .
Check. We leave the check to you.
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Solve x ( x + 2 ) 5 = 0 by using the Quadratic Formula.

x = −1 ± 6

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Solve y ( 3 y 1 ) 2 = 0 by using the Quadratic Formula.

y = 2 3 , y = 1

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Practice Key Terms 1

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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