Question:
The number of real roots of the equation $\left(x^{2}+2 x\right)^{2}-(x+1)^{2}-55=0$ is
(a) 2
(b) 1
(c) 4
(d) none of these
Solution:
(a) 2
$\left(x^{2}+2 x\right)^{2}-(x+1)^{2}-55=0$
$\Rightarrow\left(x^{2}+2 x+1-1\right)^{2}-(x+1)^{2}-55=0$
$\Rightarrow\left\{(x+1)^{2}-1\right\}^{2}-(x+1)^{2}-55=0$
$\Rightarrow\left\{(x+1)^{2}\right\}^{2}+1-3(x+1)^{2}-55=0$
$\Rightarrow\left\{(x+1)^{2}\right\}^{2}-3(x+1)^{2}-54=0$
Let $p=(x+1)^{2}$
$\Rightarrow p^{2}-3 p-54=0$
$\Rightarrow p^{2}-9 p+6 p-54=0$
$\Rightarrow(p+6)(p-9)=0$
$\Rightarrow p=9 \quad$ or $\quad p=-6$
Rejecting $p=-6$
$\Rightarrow(x+1)^{2}=9$
$\Rightarrow x^{2}+2 x-8=0$
$\Rightarrow x^{2}+4 x-2 x-8=0$
$\Rightarrow(x+4)(x-2)=0$
$\Rightarrow x=2, \quad x=-4$