Question
Solve the following quadratic equations by factorization:
$a^2x^2 - 3abx + 2b^2 = 0$
$a^2x^2 - 3abx + 2b^2 = 0$
✓
Answer
We have,
$a^2x^2 - 3abx + 2b^2 = 0$
$\Rightarrow a^2x^2 - abx - 2abx + 2b^2 = 0$
$[a^2 \times 2b^2 = 2a^2b^2 \Rightarrow 2a^2b^2 = 2ab \times ab = -2ab \times -ab\Rightarrow -3ab = -2ab - ab]$
$\Rightarrow ax(ax - b) - 2b(ax - b) = 0$
$\Rightarrow (ax - 2b)(ax - b) = 0$
$\Rightarrow ax = 2b$ or $ax = b$
$\Rightarrow\text{x}=\frac{2\text{b}}{\text{a}}$ or $\text{x}=\frac{\text{b}}{\text{a}}$
$\therefore\text{x}=\frac{\text{b}}{\text{a}}$ and $\text{x}=\frac{2\text{b}}{\text{a}}$ are the two roots of the given quadratic equation.
$a^2x^2 - 3abx + 2b^2 = 0$
$\Rightarrow a^2x^2 - abx - 2abx + 2b^2 = 0$
$[a^2 \times 2b^2 = 2a^2b^2 \Rightarrow 2a^2b^2 = 2ab \times ab = -2ab \times -ab\Rightarrow -3ab = -2ab - ab]$
$\Rightarrow ax(ax - b) - 2b(ax - b) = 0$
$\Rightarrow (ax - 2b)(ax - b) = 0$
$\Rightarrow ax = 2b$ or $ax = b$
$\Rightarrow\text{x}=\frac{2\text{b}}{\text{a}}$ or $\text{x}=\frac{\text{b}}{\text{a}}$
$\therefore\text{x}=\frac{\text{b}}{\text{a}}$ and $\text{x}=\frac{2\text{b}}{\text{a}}$ are the two roots of the given quadratic equation.
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