How do you solve $x^{2} - 8 x + 16 = 0$ $x^2−8x+16=0$ using the quadratic formula?

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How do you solve $x^{2} - 8 x + 16 = 0$ $x^2−8x+16=0$ using the quadratic formula?

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Answer 1 · 2017-02-22T00:58:19

$x = 4$ $x=4$

Explanation:

The quadratic formula is defined by:
$x = \frac{- b ((\frac{+}{-}) \sqrt{b^{2} - 4 a c})}{2 a}$ $x=(-b((+/-)sqrt(b^2-4ac)))/(2a)$
where $a$ $a$ is the coefficient in front of $x^{2}$ $x^2$ , $b$ $b$ is the coefficient in front of $x$ $x$ , and c is the last coefficient.

Plugging in 1 for a, -8 for b, and 16 for c:

$x = \frac{8 ((\frac{+}{-}) \sqrt{64 - 4 (1) (16)})}{2 (1)}$ $x=(8((+/-)sqrt(64-4(1)(16))))/(2(1))$

Simplify:

$x = \frac{8 ((\frac{+}{-}) \sqrt{0})}{2 (1)}$ $x=(8((+/-)sqrt(0)))/(2(1))$

$\sqrt{0} = 0$ $sqrt0=0$ and the positive and negative of 0 is just 0. So it becomes:

$x = \frac{8}{2} = 4$ $x=8/2=4$

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How do you solve $x^{2} - 8 x + 16 = 0$ $x^2−8x+16=0$ using the quadratic formula?

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Explanation:

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How do you solve $x^{2} - 8 x + 16 = 0$ $x^2−8x+16=0$ using the quadratic formula?