Question
Solve the following quadratic equations by factorization:
$\text{x}^2+\Big(\text{a}+\frac{1}{\text{a}}\Big)\text{x}+1=0$
$\text{x}^2+\Big(\text{a}+\frac{1}{\text{a}}\Big)\text{x}+1=0$
✓
Answer
We have been given
$\text{x}^2+\Big(\text{a}+\frac{1}{\text{a}}\Big)\text{x}+1=0$
Therefore,
$\text{x}^2+\text{ax}+\frac{1}{\text{a}}\text{x}+1=0$
$\text{x}(\text{x}+\text{a})+\frac{1}{\text{a}}(\text{x}+\text{a})=0$
$\Big(\text{x}+\frac{1}{\text{a}}\Big)(\text{x}+\text{a})=0$
Therefore,
$\text{x}+\frac{1}{\text{a}}=0$
$\text{x}=-\frac{1}{\text{a}}$
or, $\text{x}+\text{a}=0$
$\text{x}=-\text{a}$
Hence, $\text{x}=-\frac{1}{\text{a}}$ or $\text{x}=-\text{a}$
$\text{x}^2+\Big(\text{a}+\frac{1}{\text{a}}\Big)\text{x}+1=0$
Therefore,
$\text{x}^2+\text{ax}+\frac{1}{\text{a}}\text{x}+1=0$
$\text{x}(\text{x}+\text{a})+\frac{1}{\text{a}}(\text{x}+\text{a})=0$
$\Big(\text{x}+\frac{1}{\text{a}}\Big)(\text{x}+\text{a})=0$
Therefore,
$\text{x}+\frac{1}{\text{a}}=0$
$\text{x}=-\frac{1}{\text{a}}$
or, $\text{x}+\text{a}=0$
$\text{x}=-\text{a}$
Hence, $\text{x}=-\frac{1}{\text{a}}$ or $\text{x}=-\text{a}$
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