Question

plz solve question no. 65 of quadratic equation by factorisation method

Answer 1

$Given : 3(\frac{3x - 1}{2x + 3}) - 2(\frac{2x + 3}{3x - 1}) = 5$

$= > \frac{9x - 1}{2x + 3} - \frac{4x + 6}{3x - 1} = 5$

$= > \frac{(9x - 3)(3x - 1) - (4x + 6)(2x + 3)}{(2x + 3)(3x - 1)} = 5$

$= > \frac{27x^2 - 18x + 3 - 8x^2 - 24x - 18}{6x^2 + 7x - 3} = 5$

$= > \frac{19x^2 - 42x - 15}{6x^2 + 7x - 3} = 5$

$= > 19x^2 - 42x - 15 = 5(6x^2 + 7x - 3)$

⇒ 19x^2 - 42x - 15 = 30x^2 + 35x - 15

⇒ 19x^2 - 42x - 15 - 30x^2 - 35x + 15 = 0

⇒ 19x^2 - 77x - 30x^2 = 0

⇒ -11x^2 - 77x = 0

⇒ -x^2 - 7x = 0

⇒ x(x + 7) = 0

⇒ x = 0, -7.

Hope this helps!