How do you find the zeros, real and imaginary, of $y = - x^{2} - 12 x + 11$ $y=-x^2-12x+11$ using the quadratic formula?

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How do you find the zeros, real and imaginary, of $y = - x^{2} - 12 x + 11$ $y=-x^2-12x+11$ using the quadratic formula?

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Answer 1 · 2018-01-08T12:58:32

See a solution process below:

Explanation:

The quadratic formula states:

For $a x^{2} + b x + c = 0$ $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 \cdot a}$ $x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$- 1$ $color(red)(-1)$ for $a$ $color(red)(a)$

$- 12$ $color(blue)(-12)$ for $b$ $color(blue)(b)$

$11$ $color(green)(11)$ for $c$ $color(green)(c)$ gives:

$x = \frac{- - 12 \pm \sqrt{{- 12}^{2} - (4 \cdot - 1 \cdot 11)}}{2 \cdot - 1}$ $x = (-color(blue)(-12) +- sqrt(color(blue)(-12)^2 - (4 * color(red)(-1) * color(green)(11))))/(2 * color(red)(-1))$

$x = \frac{12 \pm \sqrt{144 - (- 44)}}{- 2}$ $x = (12 +- sqrt(144 - (-44)))/(-2)$

$x = \frac{12 \pm \sqrt{144 + 44}}{- 2}$ $x = (12 +- sqrt(144 + 44))/(-2)$

$x = \frac{12 \pm \sqrt{188}}{- 2}$ $x = (12 +- sqrt(188))/(-2)$

$x = \frac{12 \pm \sqrt{4 \times 47}}{- 2}$ $x = (12 +- sqrt(4 xx 47))/(-2)$

$x = \frac{12 \pm \sqrt{4} \sqrt{47}}{- 2}$ $x = (12 +- sqrt(4)sqrt(47))/(-2)$

$x = \frac{12 \pm 2 \sqrt{47}}{- 2}$ $x = (12 +- 2sqrt(47))/(-2)$

$x = - 6 \pm \sqrt{47}$ $x = -6 +- sqrt(47)$

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How do you find the zeros, real and imaginary, of $y = - x^{2} - 12 x + 11$ $y=-x^2-12x+11$ using the quadratic formula?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of y=−x2−12x+11y=-x^2-12x+11 using the quadratic formula?

1 Answer

Explanation:

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Impact of this question

How do you find the zeros, real and imaginary, of $y = - x^{2} - 12 x + 11$ $y=-x^2-12x+11$ using the quadratic formula?