Question

$2\left(\frac{2x+3}{x-3}\right)-25\left(\frac{x-3}{2x+3}\right)=5$. Solve for $x$. Write nature of roots.

Answer 1

$\left(\frac{2x+3}{x-3}\right)-25\left(\frac{x-3}{2x+3}\right)=5$

$Let\frac{2x+3}{x-3}=y$

$2y-25\times \frac{1}{y}=5$

$2y^2-25=5y$

$2y^2-25-25=0$

$2y^2+5y-1y-25=0$

$y\left(2y+5\right)-5\left(2y+5\right)=0$

$\left(2y+5\right)\left(y-5\right)=0$

$y=\frac{-5}{2},5$

$ify=\frac{-5}{2}$

$\frac{2x+3}{x-3}=\frac{-5}{2}$

$4x+6=-5x+5$

$9x=9$

$x=1$

$ify=5$

$\frac{2x+3}{x-3}=5$

$2x+3=5x-15$$

$3x=18$

$x=6$

Therefore the roots are real and distinct