How do you find the roots, real and imaginary, of $y= -x^2 - 12x + 4$ using the quadratic formula?

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Answer 1 · 2017-10-02T12:13:08

See a solution process below:

To find the roots of the equation set $y$ equal to $0$ and solve for $x$ .

$-x^2 - 12x + 4 = 0$

Now, we can use the quadratic equation to solve this problem:

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

$x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

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

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

$color(green)(4)$ for $color(green)(c)$ gives:

$x = (-color(blue)((-12)) +- sqrt(color(blue)((-12))^2 - (4 * color(red)(-1) * color(green)(4))))/(2 * color(red)(-1))$

$x = (12 +- sqrt(144 - (-16)))/(-2)$

$x = (12)/-2 +- (sqrt(144 + 16))/(-2)$

$x = -6 +- (sqrt(160))/(-2)$

$x = -6 - (sqrt(16 * 10))/(-2)$ and $x = -6 + (sqrt(16 * 10))/(-2)$

$x = -6 - (sqrt(16)sqrt(10))/(-2)$ and $x = -6 + (sqrt(16)sqrt(10))/(-2)$

$x = -6 - (4sqrt(10))/(-2)$ and $x = -6 + (4sqrt(10))/(-2)$

$x = -6 - (-2sqrt(10))$ and $x = -6 + (-2sqrt(10))$

$x = -6 + 2sqrt(10)$ and $x = -6 - 2sqrt(10)$

How do you find the roots, real and imaginary, of y= -x^2 - 12x + 4 y= -x^2 - 12x + 4 using the quadratic formula?