Question

Solve each of the following quadratic equations:
$\frac{3x-4}{7}+\frac{7}{3x-4}=\frac{5}{2},x\ne \frac{4}{3}$

Answer 1

Given:

$\frac{3x-4}{7}+\frac{7}{3x-4}=\frac{5}{2},x\ne \frac{4}{3}$

$\Rightarrow \frac{(3x-4{)}^2+49}{7(3x-4)}=\frac{5}{2},x\ne \frac{4}{3}$

$\Rightarrow \frac{9x^2-24x+16+49}{21x-28}=\frac{5}{2},x\ne \frac{4}{3}$, [Since, $(a-b{)}^2=a^2-2ab+b^2$]

$\Rightarrow \frac{9x^2-24x+65}{21x-28}=\frac{5}{2},x\ne \frac{4}{3}$

$\Rightarrow 2(9x^2-24x+65)=5(21x-28)$

$\Rightarrow 18x^2-48x+130=105x-240$

$\Rightarrow 18x^2-48x+130-105x+240=0$

$\Rightarrow 18x^2-153x+270=0$

$\Rightarrow 9(2x^2-17x+30)=0$

$\Rightarrow 2x^2-17x+30=0$

$\Rightarrow 2x^2-12x-5x+30=0$

$\Rightarrow 2x(x-6)-5(x-6)=0$

$\Rightarrow (x-6)(2x-5)=0$

$\Rightarrow x-6=0$ or $(2x-5)=0$

$\Rightarrow x=6$ or $x=\frac{5}{2}$

Hence, $6$ and $\frac{5}{2}$ are the roots of the given equation.