Question

If $|4x^2-16x+16|-7|x-2|=-3$, then the value(s) of $x$ is/are

• A. $1$
• B. $3$
• C. $\frac{11}{4}$
• D. $\frac{5}{4}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $1$
B $3$
C $\frac{11}{4}$
D $\frac{5}{4}$

$|4x^2-16x+16|-7|x-2|=-3\Rightarrow 4|x^2-4x+4|-7|x-2|+3=0$
$\Rightarrow 4|(x-2{)}^2|-7|x-2|+3=0\Rightarrow 4|(x-2){|}^2-7|x-2|+3=0(\because |x^2|=|x{|}^2)$

Let $|x-2|=t$
$\Rightarrow 4t^2-7t+3=0$
$\Rightarrow 4t^2-4t-3t+3=0$
$\Rightarrow (t-1)(4t-3)=0$
$\Rightarrow t=1\text{or}t=\frac{3}{4}$
$\Rightarrow |x-2|=1\text{or}|x-2|=\frac{3}{4}$
$\Rightarrow x-2=\pm 1\text{or}x-2=\pm \frac{3}{4}$
$\Rightarrow x=1,3\text{or}x=\frac{5}{4},\frac{11}{4}$
$\therefore x=1,\frac{5}{4},\frac{11}{4},3$