Question
Solve the following quadratic equation by factorization:
$a^2b^2x^2 + b^2x - a^2x - 1 = 0$
$a^2b^2x^2 + b^2x - a^2x - 1 = 0$
✓
Answer
We have
$a^2b^2x^2 + b^2x - a^2x - 1 = 0$
$[-1 \times a^2b^2 = -a^2b^2 \Rightarrow -a^2b^2 = -a^2 \times b^2 = -a^2 \times b^2]$
$\Rightarrow a^2b^2x^2 + b^2x - a^2x - 1 = 0$
$\Rightarrow b^2x(a^2x + 1) - 1(a^2x + 1) = 0$
$\Rightarrow (a^2x + 1)(b^2x - 1) = 0$
$\Rightarrow a^2x + 1 = 0$ or $b^2x - 1 = 0$
$\Rightarrow\text{x}= -\frac{1}{\text{a}^2}$ and $\text{x}= \frac{1}{\text{b}^2}$ are the two root of the given equation.
$a^2b^2x^2 + b^2x - a^2x - 1 = 0$
$[-1 \times a^2b^2 = -a^2b^2 \Rightarrow -a^2b^2 = -a^2 \times b^2 = -a^2 \times b^2]$
$\Rightarrow a^2b^2x^2 + b^2x - a^2x - 1 = 0$
$\Rightarrow b^2x(a^2x + 1) - 1(a^2x + 1) = 0$
$\Rightarrow (a^2x + 1)(b^2x - 1) = 0$
$\Rightarrow a^2x + 1 = 0$ or $b^2x - 1 = 0$
$\Rightarrow\text{x}= -\frac{1}{\text{a}^2}$ and $\text{x}= \frac{1}{\text{b}^2}$ are the two root of the given equation.
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