Question

If $|x-3||x+1|=12,$ then $x=$

• A. $-3$
• B. $5$
• C. $3$
• D. $-5$

Answer 1

Answer: (A), (B)

$5$

$|x-3||x+1|=12$
$\Rightarrow |x^2-2x-3|=12$

$\text{Case}1:x^2-2x-3\ge 0\text{i.e.,}x\in (-\infty ,-1]\cup [3,\infty )$
Then $x^2-2x-3=12$
$\Rightarrow x^2-2x-15=0$
$\Rightarrow (x-5)(x+3)=0$
$\Rightarrow x=-3,5$

$\text{Case}2:x^2-2x-3<0\text{i.e.,}x\in (-1,3)$
Then $x^2-2x-3=-12$
$\Rightarrow x^2-2x+9=0$
$\Rightarrow (x-1{)}^2+8=0$ (Not possible)