Question

The solution of the equation $|x+1{|}^2-2|x+2|-26=0$ is

• A. $\pm 7$
• B. $-7-\sqrt{29}$
• C. $\sqrt{29}$
• D. $-7$

Answer 1

Answer: (C), (D)

$-7$
$\sqrt{29}$

${|x+1|}^2-2|x+2|-26=0$

Let us consider intervals for values of $x$.

Since the leading term is squared, it will always be positive.

Now, if $x<-2$

$|x+2|=-(x+2)$

& if $x>,-2$

$|x+2|=(x+2)$

$\Rightarrow {(x+1)}^2+2(x+2)-26=0forx<-2$

$\Rightarrow x^2+2x+1+2x+4-26=0$

$\Rightarrow x^2+4x-21=0$

$\Rightarrow x=\frac{-4\pm \sqrt{{\left(4\right)}^2-4\times -21}}{2}=\frac{-4\pm \sqrt{16+84}}{2}=\frac{-4\pm 10}{2}=-7,3$

Since the accepted values of $x$ must be less than $-2$, the value which satisfy is $-7$.

Now, when $x\ge -2$

${(x+1)}^2-2(x+2)-26=0$

$x^2+2x+1-2x-4-26=0$

$x^2-29=0\Rightarrow x=\pm \sqrt{29}$

Since the acceptable value must be greater than $-2$, the answer would be $\sqrt{29}$

$\therefore$ Values of $x=-7$ and $\sqrt{29}$