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