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

Question

Ethan Adkins · Accepted Answer

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

There are no imaginary roots.

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.

Eva Abramson · Answer

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

Julia Adams · Answer

undefined To find the roots of the quadratic equation $y = -x^2 + 32x - 16$, you can use the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a = -1$, $b = 32$, and $c = -16$. Substituting these values into the formula:

$x = \frac{{-32 \pm \sqrt{{32^2 - 4(-1)(-16)}}}}{{2(-1)}}$

Simplify under the square root:

$x = \frac{{-32 \pm \sqrt{{1024 - 64}}}}{{-2}}$

$x = \frac{{-32 \pm \sqrt{{960}}}}{{-2}}$

$x = \frac{{-32 \pm 4\sqrt{{60}}}}{{-2}}$

Now, you have:

$x_1 = \frac{{-32 + 4\sqrt{{60}}}}{-2}$

$x_2 = \frac{{-32 - 4\sqrt{{60}}}}{-2}$

These are the roots of the quadratic equation.