What are the solutions to the equation (x^2-36)^2+(x^2+4x-12)^2=0 ?

The solutions to the equation (x^2-36)^2+(x^2+4x-12)^2=0 are x=-6,x=4+2i,x=4-2i

Find the zeros of (x^2-36)^2+(x^2+4x-12)^2=0

The zeros of (x^2-36)^2+(x^2+4x-12)^2=0 are x=-6,x=4+2i,x=4-2i

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(x^{2} - 36)^{2} + (x^{2} + 4 x - 12)^{2} = 0

x = - 6, x = 4 + 2 i, x = 4 - 2 i

Solution steps

(x^{2} - 36)^{2} + (x^{2} + 4 x - 12)^{2} = 0

Factor

(x^{2} - 36)^{2} + (x^{2} + 4 x - 12)^{2} : 2 (x + 6)^{2} (x^{2} - 8 x + 20)

2 (x + 6)^{2} (x^{2} - 8 x + 20) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 6 = 0 or x^{2} - 8 x + 20 = 0

Solve

x + 6 = 0 : x = - 6

Solve

x^{2} - 8 x + 20 = 0 : x = 4 + 2 i, x = 4 - 2 i

The solutions are

x = - 6, x = 4 + 2 i, x = 4 - 2 i

4 x^{3} + 6 x^{2} - 72 x - 3 = 0

2 x^{3} - 16 x^{2} - 40 x = 0

z^{5} + i = 0

x^{3} - 2 x - 5

3 (2 - x)^{3} - 2 (2 - x)^{2} - 19 (2 - x) - 6 = 0

What are the solutions to the equation (x^2-36)^2+(x^2+4x-12)^2=0 ?
The solutions to the equation (x^2-36)^2+(x^2+4x-12)^2=0 are x=-6,x=4+2i,x=4-2i
Find the zeros of (x^2-36)^2+(x^2+4x-12)^2=0
The zeros of (x^2-36)^2+(x^2+4x-12)^2=0 are x=-6,x=4+2i,x=4-2i