Question

Solve the following equation:
$\frac{(4x+1{)}^2+(2x+3{)}^2}{4x^2+12x+9}=\frac{61}{36}$

Answer 1

$\frac{{(4x+1)}^2+{(2x+3)}^2}{4x^2+12x+9}=\frac{61}{36}$

$\Rightarrow \frac{{(4x+1)}^2+{(2x+3)}^2}{{\left(2x\right)}^2+2\times 2x\times 3+9}=\frac{61}{36}$

$\Rightarrow \frac{{(4x+1)}^2}{{\left(2x\right)}^2+2\times 2x\times 3+9}+\frac{{(2x+3)}^2}{{\left(2x\right)}^2+2\times 2x\times 3+9}=\frac{61}{36}$

$\Rightarrow \frac{{(4x+1)}^2}{{(2x+3)}^2}+\frac{{(2x+3)}^2}{{(2x+3)}^2}=\frac{61}{36}$

$\Rightarrow \frac{{(4x+1)}^2}{{(2x+3)}^2}+1=\frac{61}{36}$

$\Rightarrow \frac{{(4x+1)}^2}{{(2x+3)}^2}=\frac{61}{36}-1$

$\Rightarrow \frac{{(4x+1)}^2}{{(2x+3)}^2}=\frac{61-36}{36}=\frac{25}{36}$

$\Rightarrow \frac{4x+1}{2x+3}=\sqrt{\frac{25}{36}}=\pm \frac{5}{6}$

$\Rightarrow \frac{4x+1}{2x+3}=\pm \frac{5}{6}$

$\Rightarrow 24x+6=10x+15$ or $24x+6=-10x-15$

$\Rightarrow 24x-10x=15-6$ or $24x+10x=-15-6$

$\Rightarrow 14x=9$ or $34x=-25$

$\therefore x=\frac{9}{14},\frac{-25}{34}$