Question
Solve the following quadratic equations by factorization:
$9x^2 - 6b^2x - (a^4 - b^4) = 0$
$9x^2 - 6b^2x - (a^4 - b^4) = 0$
✓
Answer
$9x^2 - 6b^2x - (a^4 - b^4) = 0$
$\Rightarrow 9x^2 - 6b^2x - (a^2 - b^2)(a^2 + b^2) = 0$
$\Rightarrow 9x^2 + 3(a^2 - b^2)x - 3 (a^2 + b^2)x - (a^2 - b^2)(a^2 + b^2) = 0$
$\Rightarrow 3x[3x + (a^2 - b^2)] - (a^2 + b^2)[3x + (a^2 - b^2)] = 0$
$\Rightarrow [3x + (a^2 - b^2)] [3x - (a^2 + b^2)] = 0$
$\Rightarrow 3x + (a^2 - b^2) = 0 or 3x - (a^2 - b^2) = 0$
$3x = b^2 - a^2 or 3x = a^2 + b^2$
$\Rightarrow\text{x}=\frac{\text{b}^2 -\text{a}^2}{3}$ or $\text{x}=\frac{\text{a}^2 +\text{b}^2}{3}$
$\Rightarrow\text{x}=\frac{\text{a}^2 -\text{b}^2}{3}$ or $\text{x}=\frac{\text{b}^2 -\text{a}^2}{3}$
Hence, the factors are $\frac{\text{a}^2 -\text{b}^2}{3}$ and $\frac{\text{b}^2 -\text{a}^2}{3}$
$\Rightarrow 9x^2 - 6b^2x - (a^2 - b^2)(a^2 + b^2) = 0$
$\Rightarrow 9x^2 + 3(a^2 - b^2)x - 3 (a^2 + b^2)x - (a^2 - b^2)(a^2 + b^2) = 0$
$\Rightarrow 3x[3x + (a^2 - b^2)] - (a^2 + b^2)[3x + (a^2 - b^2)] = 0$
$\Rightarrow [3x + (a^2 - b^2)] [3x - (a^2 + b^2)] = 0$
$\Rightarrow 3x + (a^2 - b^2) = 0 or 3x - (a^2 - b^2) = 0$
$3x = b^2 - a^2 or 3x = a^2 + b^2$
$\Rightarrow\text{x}=\frac{\text{b}^2 -\text{a}^2}{3}$ or $\text{x}=\frac{\text{a}^2 +\text{b}^2}{3}$
$\Rightarrow\text{x}=\frac{\text{a}^2 -\text{b}^2}{3}$ or $\text{x}=\frac{\text{b}^2 -\text{a}^2}{3}$
Hence, the factors are $\frac{\text{a}^2 -\text{b}^2}{3}$ and $\frac{\text{b}^2 -\text{a}^2}{3}$
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