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

Question

Hazel Anglin · Accepted 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#.

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.

Isaiah Anthony · Answer

4 and 8

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

Julia Adams · Answer

undefined The quadratic formula is used to find the roots of a quadratic equation in the form of ax^2 + bx + c = 0. The roots are given by the formula x = (-b ± √(b^2 - 4ac)) / (2a). In this case, the equation is y = -x^2 + 12x - 32, which can be rewritten as -x^2 + 12x - 32 = 0. Comparing this to the standard form, we have a = -1, b = 12, and c = -32. Plugging these values into the formula gives x = (-(12) ± √((12)^2 - 4(-1)(-32))) / (2(-1)), which simplifies to x = (-(12) ± √(144 - 128)) / (-2), x = (-(12) ± √16) / (-2), x = (-(12) ± 4) / (-2), x = (8 or 4). Thus, the roots are x = 8 and x = 4.