Question

If $\{x^2+(x+1{)}^2{\}}^{-\frac{1}{2}}=(x+2{)}^{-1},$ then $x=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

${\{x^2+(x+1{)}^2\}}^{-1/2}={(x+2)}^{-1}$

Squaring on both sides

${\{x^2+(x+1{)}^2\}}^{-1}={\left\{\right(x+2{)}^{-1}\}}^2$

${\{x^2+(x+1{)}^2\}}^{-1}={\left\{\right(x+2{)}^2\}}^{-1}$

$\Rightarrow (x^2+x^2+1+2x)=x^2+4x+4$

$\Rightarrow x^2=2x+3$

$\Rightarrow x^2-2x-3=0$

$\Rightarrow x^2-3x+x-3=0$

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

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

$\Rightarrow x=-1$ or $x=3$