How do you find the roots, real and imaginary, of #y=-x^2 +32x -16# using the quadratic formula?

2 Answers
Oct 7, 2017

Answer:

#x=16+sqrt(240)=31.49# OR #x=16-sqrt(240)=.51#

There are no imaginary roots.

Explanation:

Quadratic formula:

For #y=ax^2+bx+c#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

In your problem, #a=-1, b=32, c=-16#

So, #x=(-32+-sqrt(32^2-4(-1)(-16)))/(2(-1))#

Simplifying:

#x=(-32+-sqrt(1024-64))/(-2)#

#x=(-32+-sqrt(960))/(-2)#

#x=(-32)/-2+-sqrt(4*240)/(-2)#

#x=16+-(2sqrt(240))/(-2)#

So #x=16+sqrt(240)=31.49# OR #x=16-sqrt(240)=.51#

There are no imaginary roots.

Oct 7, 2017

Answer:

Roots are #16-4sqrt15# and #16+4sqrt15#

Explanation:

According to quadratic formula, the roots of #y=ax^2+bx+c# are

#(-b+-sqrt(b^2-4ac))/(2a)#

Hence roots of #y=-x^2+32x-16# are

#(-32+-sqrt(32^2-4xx(-1)(-16)))/(2xx(-1))#

= #(-32+-sqrt(1024-64))/(-2)#

= #16+-sqrt960/2#

= #16+-(8sqrt15)/2#

= #16+-4sqrt15#