find the solutions. x^3-12x^2+52x=0?

Question

*****there are THREE ANSWERS total******
pre cal problem. please help.

Answer 1

First: factor > x(x^2 - 12x + 52) = 0
Sec: use the Quadratic Formula, which is...
x = [-b +/- √(b^2 - 4ac)] / 2a

*Substitute the terms with the corresponding variables...
a = 1 > b = -12 > c = 52

x = [-(-12) +/- √((-12)^2 - 4(1)(52))] / 2(1)
x = [12 +/- √((-12)(-12) - 4(52))] / 2
x = [12 +/- √(144 - 208)] / 2
x = [12 +/- √(-64)] / 2

*A negative number can't be in a square root - the negative becomes an imaginary number represented as (i). Place it in front of the radical.

x = [12 +/- i√(64)] / 2
x = [12 +/- 8i] / 2
x = 12/2 +/- 8i/2
x = 6 +/- 4i

Answer 2

x^3-12x^2+52x=0
x(x^2 -12x+52)=0
using the quadratic formula since the equation does not factorize
x=[12+sqrt(-64)]/2 and [12-sqrt(-64)]/2
x=6+4i and 6-4i
The solutions are x=0, x=6+4i, x=6-4i where as usual i^2=(-1)

Answer 3

x^3 + 9x^2 - 52x = 0 ingredient out an x x(x^2 + 9x - fifty two) = 0 So, x = 0 or (x^2 + 9x - fifty two) = 0 ingredient (x^2 + 9x - fifty two) (x +___ )(x +___) ---> -fifty two = -26*2 = 26*-2 = -thirteen*4 = thirteen*-4 thirteen and -4 are the only words that upload to make 9, yet multiply to make -fifty two (x + thirteen)(x - 4) = 0 (x + thirteen) = 0 OR (x - 4) = 0 So, x = -thirteen, x = 4 or x = 0

Answer 4

x^3-12x^2+52x=0
x(x^2-12x+52)=0
x=0 OR x^2-12x+52=0
x=[12+/- sqrt(144-208)]/2
=[12+/- sqrt(-64)]/2
=6+/-4i. ANS.

Answer 5

x(x^2-12x+52)

solutions are 0 and your quadratic formula answers...do them

Answer 6

x(x^2 - 12x + 52) = 0

x(x^2 - 2x*6 + 52) = 0

x(x^2 - 2x*6 + 6^2 - 6^2 + 52) = 0

x [(x - 6)^2 - 36 + 52] = 0

x[(x - 6)^2 + 16] = 0

x[(x - 6)^2 - (4i)^2] = 0

x(x - 6 - 4i)(x - 6 + 4i) = 0

x1 = 0
x2 = 6 + 4i
x3 = 6 - 4i

Answer 7

Factor X
and solve it urself