Question

If $2|x-3{|}^2-|x-1|=8$, then $x\in$

• A. $\{1,\frac{9}{2}\}$
• B. $\{\frac{9}{2},\frac{11}{2}\}$
• C. $\{1,\frac{9}{2},\frac{11}{2}\}$
• D. $\{1,\frac{11}{2}\}$

Answer 1

The correct option is D $\{1,\frac{11}{2}\}$

$2|x-3{|}^2-|x-1|=8$
$\Rightarrow 2(x^2-6x+9)-|x-1|-8=0$
$\Rightarrow 2(x^2-6x+5)-|x-1|=0$
$\Rightarrow 2(x-1)(x-5)-|x-1|=0$

When $x\ge 1$
$(x-1)\left[2\right(x-5)-1]=0$
$\Rightarrow x=1,\frac{11}{2}$

When $x<1$
$(x-1)\left[2\right(x-5)+1]=0$
$\Rightarrow x=1,\frac{9}{2}$ which is absurd.

$\therefore x=1,\frac{11}{2}$