How do you find the zeros, real and imaginary, of $y=-x^2 -16x +8$ using the quadratic formula?

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Answer 1 · 2017-12-23T04:29:45

The zeros of the equation are $8 +- 6sqrt(2)$ .

$y = -x^2 - 16x + 8$

We can solve for any zero(s) of a function with the quadratic formula, stated here: $x = (-b+- sqrt(b^2 - 4ac))/(2a)$

Our equation is written in standard form, or $y = ax^2 + bx + c$ .

We know that $a = -1$ , $b = -16$ , and $c = 8$ , so we plug in those values into the quadratic formula and solve for $x$ :

$x = (-(-16) +- sqrt((-16)^2 - 4(-1)(8)))/(2(-1))$

$x = (16 +- sqrt(256 - 4(-8)))/-2$

$x = (16 +- sqrt(256+32))/2$

$x = (16 +- sqrt(288))/2$

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

$x = 8 +- 6sqrt(2)$

The zeros of the equation are $8 +- 6sqrt(2)$ .

Hope this helps!

How do you find the zeros, real and imaginary, of y=-x^2 -16x +8y=-x^2 -16x +8 using the quadratic formula?