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!