Question

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

Answer 1

Consider the given equation,

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

$\frac{16x^2+1+8x+4x^2+9+12x}{4x^2+12x+9}=\frac{61}{36}$

$\frac{20x^2+20x+10}{4x^2+12x+9}=\frac{61}{36}$

$720x^2+720x+360=244x^2+732x+549$

$476x^2-12x-189=0$

We know that

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Therefore,

$x=\frac{-(-12)\pm \sqrt{{(-12)}^2-4\times 476\times (-189)}}{2\times 476}$

$x=\frac{12\pm \sqrt{144+359856}}{952}$

$x=\frac{12\pm \sqrt{360,000}}{952}$

$x=\frac{12\pm 600}{952}$

$x=\frac{12+600}{952},\frac{12-600}{952}$

$x=\frac{612}{952},\frac{-588}{952}$

Hence, the value of $x$ are $\frac{612}{952},\frac{-588}{952}$