Question

Please solve this quadratic equation.
Class X Quadratic Equations

Thank you in advance! <3​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\:3\bigg(\dfrac{3x - 1}{2x + 3} \bigg) - 2\bigg(\dfrac{2x + 3}{3x - 1} \bigg) = 5$

Let assume that

$\red{\rm :\longmapsto\:\dfrac{3x - 1}{2x + 3} = y}$

So, given equation can be rewritten as

$\rm :\longmapsto\:3y - \dfrac{2}{y} = 5$

$\rm :\longmapsto\:\dfrac{ {3y}^{2} - 2}{y} = 5$

$\rm :\longmapsto\: {3y}^{2} - 2 = 5y$

$\rm :\longmapsto\: {3y}^{2} - 5y - 2 =$

$\rm :\longmapsto\: {3y}^{2} - 6y + y - 2 =$

$\rm :\longmapsto\:3y(y - 2) + 1(y - 2) = 0$

$\rm :\longmapsto\:(y - 2)(3y + 1) = 0$

$\rm :\longmapsto\:y - 2 = 0 \: \: \: or \: \: \: 3y + 1 = 0$

$\bf\implies \:y = 2 \: \: \: or \: \: \: y = - \dfrac{1}{3}$

Now, Consider

$\rm :\longmapsto\:y = 2$

$\red{\rm :\longmapsto\:\dfrac{3x - 1}{2x + 3} = 2}$

$\rm :\longmapsto\:3x - 1 = 4x + 6$

$\rm :\longmapsto\:3x - 4x = 1 + 6$

$\rm :\longmapsto\: - x = 7$

$\bf\implies \:x = - 7$

Now, Consider

$\rm :\longmapsto\:y = - \dfrac{1}{3}$

$\red{\rm :\longmapsto\:\dfrac{3x - 1}{2x + 3} = - \dfrac{1}{3} }$

$\rm :\longmapsto\:9x - 3 = - 2x - 3$

$\rm :\longmapsto\:9x + 2x = 0$

$\rm :\longmapsto\:11x = 0$

$\bf\implies \:x = 0$

Question

If given equation is like

$\rm :\longmapsto\:3\bigg(\dfrac{3x - 1}{2x + 3} \bigg) + 2\bigg(\dfrac{2x + 3}{3x - 1} \bigg) = 5$

Let assume that

$\red{\rm :\longmapsto\:\dfrac{3x - 1}{2x + 3} = y}$

The given equation can be rewritten as

$\rm :\longmapsto\:3y + \dfrac{2}{y} = 5$

$\rm :\longmapsto\:\dfrac{ {3y}^{2} + 2}{y} = 5$

$\rm :\longmapsto\: {3y}^{2} + 2 = 5y$

$\rm :\longmapsto\: {3y}^{2} - 5y + 2 = 0$

$\rm :\longmapsto\: {3y}^{2} - 3y - 2y + 2 = 0$

$\rm :\longmapsto\:3y(y - 1) - 2(y - 1) = 0$

$\rm :\longmapsto\:(y - 1)(3y - 2) = 0$

$\rm :\longmapsto\:y - 1 = 0 \: \: \: or \: \: \: 3y - 2 = 0$

$\bf\implies \:y = 1 \: \: \: or \: \: \: y = \dfrac{2}{3}$

Consider,

$\rm :\longmapsto\:y = 1$

$\red{\rm :\longmapsto\:\dfrac{3x - 1}{2x + 3} = 1}$

$\rm :\longmapsto\:3x - 1 = 2x + 3$

$\rm :\longmapsto\:3x - 2x = 1 + 3$

$\bf :\longmapsto\:x = 4$

Consider,

$\rm :\longmapsto\:y = \dfrac{2}{3}$

$\red{\rm :\longmapsto\:\dfrac{3x - 1}{2x + 3} = \dfrac{2}{3} }$

$\rm :\longmapsto\:9x - 3 = 4x + 6$

$\rm :\longmapsto\:9x - 4x= 3 + 6$

$\rm :\longmapsto\:5x= 9$

$\bf\implies \:x = \dfrac{9}{5}$

Answer 2

Here is your answer
Hope it helps you

And then solve it!!