Question

35.) Find the roots of the equation, if they exists, by applying the quadratic formula :-


$a {}^{2} b {}^{2} x {}^{2} - (4b {}^{4} - 3a {}^{4} ) - 12a {}^{2} b {}^{2} = 0$

, a # 0 and b # 0


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Answer 1

Given Equation is a^2b^2x^2 - (4b^4 - 3a^4) - 12a^2b^2 = 0

Here, a = a^2b^2, b = -4b^4 + 3a^4, c = -12a^2b^2.

∴ D = b^2 - 4ac

      = (-4b^4 + 3a^4)^2 - 4(a^2b^2)(-12a^2b^2)

      = 16b^8 + 9a^8 + 24b^4a^4.

(i)

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

$=>x=\frac{-(-4b^4 + 3a^4)+\sqrt{16b^8+9a^8+24b^4a^4}}{2(a^2b^2)}$

$=>\frac{(4b^4 - 3a^4)+ \sqrt{(4b^4+3a^4)^2}}{2a^2b^2}$

$=>\frac{(4b^4-3a^4)+{4b^4+3a^4}}{2a^2b^2}$

$=>\frac{4b^4-3a^4 + 4b^4 + 3a^4}{2a^2b^2}$

$=>x=\frac{8b^4}{2a^2b^2}$

$=>\frac{4b^2}{a^2}$

(ii)

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

$=>\frac{-(-4b^4 + 3a^4)-\sqrt{16b^8+9a^8+24b^4a^4}}{2(a^2b^2)}$

$=>\frac{4b^4-3a^4- 4b^4-3a^4}{2a^2b^2}$

$=> -\frac{6a^4}{2a^2b^2}$

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

Therefore, the roots of the equation:

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

Hope this helps!