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Solve the Following Quadratic Equations by Factorization: - Mathematics

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Question

Solve the following quadratic equations by factorization:

$9 x^2 - 6 b^2 x - \left( a^4 - b^4 \right) = 0$

Solution

$9 x^2 - 6 b^2 x - \left( a^4 - b^4 \right) = 0$

$\Rightarrow 9 x^2 - 6 b^2 x - \left( a^2 - b^2 \right)\left( a^2 + b^2 \right) = 0$

$\Rightarrow 9 x^2 + 3( a^2 - b^2 )x - 3\left( a^2 + b^2 \right)x - \left( a^2 - b^2 \right)\left( a^2 + b^2 \right) = 0$

$\Rightarrow 3x\left[ 3x + \left( a^2 - b^2 \right) \right] - \left( a^2 + b^2 \right)\left[ 3x + \left( a^2 - b^2 \right) \right] = 0$

$\Rightarrow \left[ 3x - \left( a^2 + b^2 \right) \right]\left[ 3x + \left( a^2 - b^2 \right) \right] = 0$

$\Rightarrow 3x - \left( a^2 + b^2 \right) = 0 or 3x + \left( a^2 - b^2 \right) = 0$

$\Rightarrow x = \frac{a^2 + b^2}{3} \text { or }x = - \frac{a^2 - b^2}{3}$

$\Rightarrow x = \frac{a^2 + b^2}{3} \text { or } x = \frac{b^2 - a^2}{3}$

Hence, the factors are $\frac{a^2 + b^2}{3}$ and $\frac{b^2 - a^2}{3}$.

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