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

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Answer 1 · 2018-06-02T21:52:49

$x=-1$

$x=-3$

$x=1$

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

Let $k = x^2+2x$

Substute:

$k^2-2k-3=0$

Factor:

#(k+1)(k-3)= 0

Now un-substute:

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

Factor:

$(x+1)(x+1)(x+3)(x-1)=0$

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

$x=-1$

$x=-3$

$x=1$

graph{(x+1)^2(x+3)(x-1) [-10, 10, -5, 5]}

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