Question

The number of real roots of the equation $(x^2+2x{)}^2-(x+1{)}^2-55=0$ is
(a) 2
(b) 1
(c) 4
(d) none of these

Answer 1

(a) 2

$$\begin{gathered} {\left(x^2+2x\right)}^2-{\left(x+1\right)}^2-55=0 \\ \Rightarrow {\left(x^2+2x+1-1\right)}^2-{\left(x+1\right)}^2-55=0 \\ \Rightarrow {\left\{{\left(x+1\right)}^2-1\right\}}^2-{\left(x+1\right)}^2-55=0 \\ \Rightarrow {\left\{{\left(x+1\right)}^2\right\}}^2+1-3{\left(x+1\right)}^2-55=0 \\ \Rightarrow {\left\{{\left(x+1\right)}^2\right\}}^2-3{\left(x+1\right)}^2-54=0 \\ \mathrm{Let}p={\left(x+1\right)}^2 \\ \Rightarrow p^2-3p-54=0 \\ \Rightarrow p^2-9p+6p-54=0 \\ \Rightarrow \left(p+6\right)\left(p-9\right)=0 \\ \Rightarrow p=9\mathrm{or}p=-6 \end{gathered}$$

$$\begin{gathered} \mathrm{Rejecting}\mathit{}p=-6 \\ \Rightarrow {\left(x+1\right)}^2=9 \\ \Rightarrow x^2+2x-8=0 \\ \Rightarrow x^2+4x-2x-8=0 \\ \Rightarrow \left(x+4\right)\left(x-2\right)=0 \\ \Rightarrow x=2,x=-4 \end{gathered}$$