How do you find the zeros, real and imaginary, of $y=-x^2-3x+11$ using the quadratic formula?

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Answer 1 · 2017-10-18T18:12:54

See a solution process below:

We can use the quadratic equation to solve this problem by equating the quadratic to $0$ to find the roots or the zeros:

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 $color(red)(a)$

$color(blue)(-3)$ for $color(blue)(b)$

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

$x = (-color(blue)((-3)) +- sqrt(color(blue)((-3))^2 - (4 * color(red)(-1) * color(green)(11))))/(2 * color(red)(-1))$

$x = (color(blue)((-3)) +- sqrt(9 - (-44)))/(-2)$

$x = (color(blue)((-3)) +- sqrt(9 + 44))/(-2)$

$x = (-3 +- sqrt(53))/(-2)$

$x = (3 +- sqrt(53))/2$

How do you find the zeros, real and imaginary, of y=-x^2-3x+11y=-x^2-3x+11 using the quadratic formula?