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Algebra 1

Course: Algebra 1 > Unit 14

Lesson 6: The quadratic formula
Math
Algebra 1
Quadratic functions & equations
The quadratic formula
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Quadratic formula review

Google Classroom
The quadratic formula allows us to solve any quadratic equation that's in the form ax^2 + bx + c = 0. This article reviews how to apply the formula.

What is the quadratic formula?

The quadratic formula says that
x, equals, start fraction, minus, start color #e07d10, b, end color #e07d10, plus minus, square root of, start color #e07d10, b, end color #e07d10, squared, minus, 4, start color #7854ab, a, end color #7854ab, start color #e84d39, c, end color #e84d39, end square root, divided by, 2, start color #7854ab, a, end color #7854ab, end fraction
for any quadratic equation like:
start color #7854ab, a, end color #7854ab, x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #e84d39, c, end color #e84d39, equals, 0

Example

We're given an equation and asked to solve for q:
0, equals, minus, 7, q, squared, plus, 2, q, plus, 9
This equation is already in the form a, x, squared, plus, b, x, plus, c, equals, 0, so we can apply the quadratic formula where a, equals, minus, 7, comma, b, equals, 2, comma, c, equals, 9:
q=−b±b2−4ac2aq=−2±22−4(−7)(9)2(−7)q=−2±4+252−14q=−2±256−14q=−2±16−14q=−2+16−14  ,  q=−2−16−14q=−1            ,  q=97\begin{aligned} q &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\\ q &= \dfrac{-2 \pm \sqrt{2^{2} - 4 (-7) (9)}}{2(-7)} \\\\ q &= \dfrac{-2 \pm \sqrt{4 +252}}{-14} \\\\ q &= \dfrac{-2 \pm \sqrt{256}}{-14} \\\\ q &= \dfrac{-2 \pm 16}{-14} \\\\ q &= \dfrac{-2 + 16}{-14} ~~,~~ q = \dfrac{-2 - 16}{-14} \\\\ q &= -1 ~~~~~~~~~~~~,~~ q = \dfrac{9}{7} \end{aligned}
Let's check both solutions to be sure it worked:
q, equals, minus, 1q, equals, start fraction, 9, divided by, 7, end fraction
0=−7q2+2q+90=−7(−1)2+2(−1)+90=−7(1)−2+90=−7−2+90=0\begin{aligned}0&=-7q^2+2q+9\\\\0&=-7(-1)^2+2(-1)+9 \\\\0&=-7(1)-2+9 \\\\0&=-7-2+9\\\\0&=0\end{aligned}0=−7q2+2q+90=−7(97)2+2(97)+90=−7(8149)+(187)+90=−(817)+(187)+90=−(637)+90=−9+90=0\begin{aligned}0&=-7q^2+2q+9\\\\0&=-7\left(\dfrac{9}{7}\right)^2+2\left (\dfrac{9}{7}\right)+9 \\\\0&=-7\left(\dfrac{81}{49}\right)+\left (\dfrac{18}{7}\right)+9 \\\\0&=-\left(\dfrac{81}{7}\right)+\left (\dfrac{18}{7}\right)+9 \\\\0&=-\left(\dfrac{63}{7}\right) +9 \\\\0&=-9 +9 \\\\0&=0\end{aligned}
Yep, both solutions check out.
Want to learn more about the quadratic formula? Check out this video.
Practice
Solve for x.
minus, 4, plus, x, plus, 7, x, squared, equals, 0
Choose 1 answer:

Want more practice? Check out this exercise.

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