What is the answer to (x^2+2x-3)/((x+1)^2)=0 ?

The answer to (x^2+2x-3)/((x+1)^2)=0 is x=1,x=-3

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(x^2+2x-3)/((x+1)^2)=0

\frac{x ^{2} + 2 x - 3}{( x + 1 ) ^{2}} = 0

x = 1, x = - 3

Solution steps

\frac{x ^{2} + 2 x - 3}{( x + 1 ) ^{2}} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Solve

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

x = 1, x = - 3

Verify Solutions

Find undefined (singularity) points:

x = - 1

Combine undefined points with solutions:

x = 1, x = - 3

\frac{2 x - 4}{x - 2} - 2 = 0

2 x - \frac{432}{x ^{2}} = 0

x, 0 = \frac{2}{x}

\frac{x - 4}{2 x + 1} = \frac{x - 9}{x}

\frac{1}{( x + 3 ) ( x + 3 )} = 1, x \neq = - 3

What is the answer to (x^2+2x-3)/((x+1)^2)=0 ?
The answer to (x^2+2x-3)/((x+1)^2)=0 is x=1,x=-3
Solve for x: (x^2+2x-3)/((x+1)^2)=0
The solution is x=1,x=-3