How do you solve $(x^2+2x)^2-2(x^2+2x)-3=0$ ?

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Answer 1 · 2016-07-14T00:20:59

$x = {-3,-1,1}$

First making $y = x^2+2x$ and substituting

$y^2-2y-3=0$

Solving for $y$ gives $y = {-1,3}$

The next step is to solve

1) $x^2+2x = -1$ giving two equal roots $x = {-1,-1}$
2) $x^2+2x = 3$ giving $x = {-3, 1}$

The solutions are

$x = {-3,-1,1}$

also we have

$(x^2 + 2 x)^2 - 2 (x^2 + 2 x) - 3=(x+3)(x+1)^2(x-1)$

How do you solve (x^2+2x)^2-2(x^2+2x)-3=0 (x^2+2x)^2-2(x^2+2x)-3=0?