On Solving the following quadratic equation by factorization, the roots are $\frac{a^{2} + b^{2}}{3}, \frac{b^{2} - a^{2}}{3}$ :
$9 x^{2} - 6 b^{2} x - (a^{4} - b^{4}) = 0$

True

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False

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Solution

The correct option is A True
$9 x^{2} - 6 b^{2} x - (a^{4} - b^{4}) = 0$
$\Rightarrow$ $9 x^{2} - 3 (2 b^{2}) x - (a^{4} - b^{4}) = 0$
$\Rightarrow$ $9 x^{2} - 3 (b^{2} - a^{2} + b^{2} + a^{2}) x + (b^{4} - a^{4}) = 0$
$\Rightarrow$ $9 x^{2} - 3 (b^{2} - a^{2}) x - 3 (b^{2} + a^{2}) x + (b^{2} + a^{2}) (b^{2} - a^{2}) = 0$
$\Rightarrow$ $3 x (3 x - b^{2} + a^{2}) - (b^{2} + a^{2}) (3 x - b^{2} + a^{2}) = 0$
$\Rightarrow$ $(3 x - b^{2} - a^{2}) (3 x - b^{2} + a^{2}) = 0$
$\Rightarrow$ $3 x - b^{2} - a^{2} = 0$ and $3 x - b^{2} + a^{2} = 0$
$\Rightarrow$ $3 x = a^{2} + b^{2}$ and $3 x = b^{2} - a^{2}$
$∴$ $x = \frac{a^{2} + b^{2}}{3}$ and $x = \frac{b^{2} - a^{2}}{3}$
$∴$ Roots given in question are correct.

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