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High School
Algebra
Algebra I
Quadratic function and equation

Lennie Carroll

Answered question

2021-02-09

Solution to quadratic equation Quadratic formula -2 Examples with solution and checking

Answer & Explanation

Velsenw

Skilled2021-02-10Added 91 answers

The quadratic formula for an equation $a x^{2} + b x + c = 0 i s g i v e n b y x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2} a$
Consider the quadratic equation $x^{2} + 5 x + 6 = 0 .$
Compare the equation with standard form $a x^{2} + b x + c = 0 .$
Here, $a = 1, b = 5 a n d c = 6.$
Evaluate the solutions of the quadratic equation as follows.
$x = \frac{- 5 \pm \sqrt{5^{2} - 4 (1) (6)}}{2} (1)$
$x = \frac{- 5 \pm \sqrt{25 - 24}}{2}$
$x = \frac{5 \pm 1}{2}$
$x = \frac{- 5 + 1}{2} or x = \frac{- 5 - 1}{2}$
$x = - 2 o r x = - 3$
Verification:
Substitute $x = - 2$ in the quadratic equation and verify the solution as follows.
${(- 2)}^{2} + 5 (- 2) + 6 = 0$
$4 - 10 + 6 = 0$
$0 = 0$ (True)
Thus, the solution $x = - 2$ is verified.
Substitute $x = - 3$ in the quadratic equation and verify the solution as follows.
${(- 3)}^{2} + 5 (- 3) + 6 = 0$
$9 - 15 + 6 = 0$
$0 = 0$ (True)
Thus, the solution $x = - 3$ is verified.

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