What are the solutions to the equation (x^2+2x-3)^2=0 ?

The solutions to the equation (x^2+2x-3)^2=0 are x=1,x=-3

Find the zeros of (x^2+2x-3)^2=0

The zeros of (x^2+2x-3)^2=0 are x=1,x=-3

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(x^{2} + 2 x - 3)^{2} = 0

x = 1, x = - 3

Solution steps

(x^{2} + 2 x - 3)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

x^{2} + 2 x - 3 = 0 : x = 1, x = - 3

The solutions are

x = 1, x = - 3

2 x^{2} - 4 x^{4} = 5

x^{3} - 5 x^{2} + 5 x - 2 = 0

- 1 + v^{3} = - 3

x (x + 3) (x - 2) = 0

x^{3} + 6 x^{2} - 36 x + 2 = 0

What are the solutions to the equation (x^2+2x-3)^2=0 ?
The solutions to the equation (x^2+2x-3)^2=0 are x=1,x=-3
Find the zeros of (x^2+2x-3)^2=0
The zeros of (x^2+2x-3)^2=0 are x=1,x=-3