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

2 Answers
May 13, 2016

Answer:

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.

May 13, 2016

Answer:

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