Question
Solve the following quadratic equations by factorization:
$ax^2 + (4a^2 - 3b)x - 12ab = 0$
$ax^2 + (4a^2 - 3b)x - 12ab = 0$
✓
Answer
We have,
$ax^2 + (4a^2 - 3b)x - 12ab = 0$
$[a \times 12ab = -12a^2b^2 = 4a^2 \times -3b]$
$\Rightarrow ax^2 + 4a^2x - 3bx + (4a \times (-3b)) = 0$
$\Rightarrow ax(x + 4a) - 3b(x + 4a) = 0$
$\Rightarrow (x + 4a)(ax - 3b) = 0$
$\Rightarrow$ $(x + 4a) = 0$ or $(ax - 3b) = 0$
$\Rightarrow$ $x = -4a$ or $\text{x}=\frac{3\text{b}}{\text{a}}$
$\therefore\text{x}=\frac{3\text{b}}{\text{a}}$ and $x = -4a$ are the two roots of the given equations.
$ax^2 + (4a^2 - 3b)x - 12ab = 0$
$[a \times 12ab = -12a^2b^2 = 4a^2 \times -3b]$
$\Rightarrow ax^2 + 4a^2x - 3bx + (4a \times (-3b)) = 0$
$\Rightarrow ax(x + 4a) - 3b(x + 4a) = 0$
$\Rightarrow (x + 4a)(ax - 3b) = 0$
$\Rightarrow$ $(x + 4a) = 0$ or $(ax - 3b) = 0$
$\Rightarrow$ $x = -4a$ or $\text{x}=\frac{3\text{b}}{\text{a}}$
$\therefore\text{x}=\frac{3\text{b}}{\text{a}}$ and $x = -4a$ are the two roots of the given equations.
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