Question

Solve each of the following quadratic equations:
$\left(\frac{4x-3}{2x+1}\right)-10\left(\frac{2x+1}{4x-3}\right)=3,x\ne \frac{-1}{2},\frac{3}{4}$

Answer 1

$\left(\frac{4x-3}{2x+1}\right)-10\left(\frac{2x+1}{4x-3}\right)=3$

$\frac{(4x-3{)}^2-10(2x+1{)}^2}{(2x+1)(4x-3)}=3$​​​​​​​

$(4x-3{)}^2-10(2x+1{)}^2=3(2x+1)(4x-3)$​​​​​​​

$16x^2+9-24x-40x^2-10-40x=3(8x^2+4x-6x-3)$​​​​​​​

$-24x^2-64x-1=24x^2-6x-9$​​​​​​​

$48x^2+58x-8=0$​​​​​​​

$24x^2+29x-4=0$​​​​​​​

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\frac{-29\pm \sqrt{{29}^2-4\times 24\times (-4)}}{2\times 24}\frac{-29\pm \sqrt{841+384}}{48}\frac{-29\pm \sqrt{1225}}{48}\frac{-29\pm 35}{48}\frac{-29+35}{48}or\frac{-29-35}{48}\frac{6}{48}or\frac{-64}{48}\frac{1}{8}or\frac{-4}{3}$

Values of x are $\frac{1}{8}or\frac{-4}{3}$