Question

Find the roots, if they exists, by applying the quadratic formula :-

$12abx {}^{2} - (9a {}^{2} - 8b {}^{2} )x - 6ab = 0$

, where a # 0 and b # 0

Answer 1

Given Equation is 12abx^2 - (9a^2 - 8b^2)x - 6ab = 0

Here, a = 12ab, b = -9a^2 + 8b^2, c = -6ab.

∴ D = b^2 - 4ac

      = (-9a^2 + 8b^2)^2 - 4(12ab)(-6ab)

      = 81a^2 + 64b^2 + 144a^2b^2.

(i)

$=>x=\frac{-b+ \sqrt{D}}{2a}$

$=>\frac{-(-9a^2+8b^2)+\sqrt{81a^2 + 64b^2 + 144a^2b^2}}{2(12ab)}$

$=>\frac{9a^2-8b^2+\sqrt{(9a + 8b)^2}}{24ab}$

$=>x=\frac{9a^2-8b^2+9a^2+8b^2}{24ab}$

$=> x=\frac{18a^2}{24ab}$

$=>\frac{18a}{24b}$

$=> \frac{3a}{4b}$

(ii)

$=>x=\frac{-b-\sqrt{D}}{2a}$

$=>\frac{-(-9a^2+8b^2)-\sqrt{81a^2+64b^2+144a^2b^2}}{2(12ab)}$

$=>\frac{9a^2-8b^2-9a^2-8b^2}{24ab}$

$=>-\frac{16b^2}{24ab}$

$=>-\frac{2b}{3a}$

Therefore, the roots of the equation:

$=>x=\boxed{\frac{3a}{4b},- \frac{2b}{3a}}$

Hope it helps!