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

2 Answers
M7chael
Oct 7, 2017

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.

Shwetank Mauria
Oct 7, 2017

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

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