Question

If $\frac{1}{5}(3x-2)-\frac{1}{3}(x+7)+1=0$; find the value of $4x^2-4x+1$.

• A. $154$
• B. $134$
• C. $124$
• D. $144$

Answer 1

Answer: (D)

$144$

$\frac{1}{5}(3x-2)-\frac{1}{3}(x+7)+1=0$
$\frac{9x-6-5x-35+15}{15}=0$
$\frac{4x-26}{15}=0$
$4x=26$
$2x=13$
So, value of
$4x^2-4x+1=13\ast 13-26+1$
$4x^2-4x+1=144$