How do you find the roots, real and imaginary, of y=-x^2 +12x -32 using the quadratic formula?

2 Answers
David G.
May 13, 2016

The roots are the points on the curve where y=0, so set -x^2+12x-32=0.

Solve using the quadratic formula as shown below.

The solutions (roots) are x=4 and x=8.

Explanation:

For a quadratic equation in standard form, ax^2+bx+c=0, we know:

y=(-b+-sqrt(b^2-4ac))/(2a) (the quadratic formula)

In this instance:

a=-1
b=12
c=-32

So:

y=(-12+-sqrt(12^2-4xx(-1)xx(-32)))/(2xx(-1))

=(-12+-sqrt(144-128))/-2=(-12+-sqrt(16))/-2=(-12+-4)/-2

=(-8)/-2=4 or (-16)/-2=8

So the roots are 4 and 8.

Nghi N.
May 13, 2016

4 and 8

Explanation:

Use the improved quadratic formula (Socratic Search)
y = -x^2 + 12x - 32 = 0
D = d^2 = b^2 - 4ac = 144 - 128 = 16 --> d = +- 4
There are 2 real roots:
x = -b/(2a) +- d/(2a) = -12/-2 +- 4/-2 = 6 +- 2

x1 = 6 + 2 = 8 and x2 = 6 - 2 = 4

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