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

The solutions to the equation (x-4)^2(x+2)=(x+2)^2(x-4) are x=4,x=-2

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

The zeros of (x-4)^2(x+2)=(x+2)^2(x-4) are x=4,x=-2

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(x - 4)^{2} (x + 2) = (x + 2)^{2} (x - 4)

x = 4, x = - 2

Solution steps

(x - 4)^{2} (x + 2) = (x + 2)^{2} (x - 4)

Move

(x + 2)^{2} (x - 4)

to the left side

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

Factor

(x - 4)^{2} (x + 2) - (x + 2)^{2} (x - 4) : - 6 (x - 4) (x + 2)

- 6 (x - 4) (x + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 4 = 0 or x + 2 = 0

Solve

x - 4 = 0 : x = 4

Solve

x + 2 = 0 : x = - 2

The solutions to the quadratic equation are:

x = 4, x = - 2

x (x^{2} + 2 x + 3) = x^{3} - 1

2 x^{2} = - 9 x - 10

15 m (m + 5) = 38 m - 20

x^{2} + 2 4^{2} = 2 6^{2}

x^{2} - 4 x - 3 = 0

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