Question
Solve the following quadratic equations by factorization:
$\text{x}-\frac{1}{\text{x}}=3,\text{x}\neq0$
$\text{x}-\frac{1}{\text{x}}=3,\text{x}\neq0$
✓
Answer
$\text{x}-\frac{1}{\text{x}}=3$
$\Rightarrow\text{x}^2-3\text{x}-1=0$
Here $\text{a}=1,\text{b}=-3,\text{c}=-1$
$\therefore\text{D}=\text{b}^2-4\text{ac}$
$=(-3)^2-4\times(1)(-1)$
$=9+4=13$
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$\text{x}={-(-3) \pm \sqrt{13} \over 2\times1}$
$\text{x}=\frac{3\pm\sqrt{13}}{2}$
$\Rightarrow\text{x}^2-3\text{x}-1=0$
Here $\text{a}=1,\text{b}=-3,\text{c}=-1$
$\therefore\text{D}=\text{b}^2-4\text{ac}$
$=(-3)^2-4\times(1)(-1)$
$=9+4=13$
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$\text{x}={-(-3) \pm \sqrt{13} \over 2\times1}$
$\text{x}=\frac{3\pm\sqrt{13}}{2}$
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