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Q.
If $\alpha, \beta$ are the roots of the quadratic equation $x^2+ax+b=0, (b\ne 0)$; then the quadratic equation whose roots are $\alpha -\frac{1}{\beta}, \beta - \frac{1}{\alpha}$ is
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Solution:
Given equation is, $x^{2}+a x+b=0,(b \neq 0)$
its roots are $\alpha$ and $\beta$.
Then, sum of roots $=\alpha+\beta=-a$ ....(i)
Product of roots $=\alpha \cdot \beta=b$ .....(ii)
Now,
$\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)=(\alpha+\beta)-\left(\frac{\alpha+\beta}{\alpha \beta}\right)$
$=-a-\frac{(-a)}{b}$ [from Eqs.(i) and (ii)]
$=-a+\frac{a}{b}=\frac{a}{b}(1-b) $
and $\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)=\alpha \beta-1-1+\frac{1}{\alpha \beta} $
$=b+\frac{1}{b}-2[\text { from Eq. (ii) }]$ ....(iv)
$=\frac{1}{b}\left(b^{2}-2 b+1\right)=\frac{1}{b}(b-1)^{2}$
$\therefore $ Required of quadratic equation whose roots are $\left(\alpha-\frac{1}{\beta}\right)$ and $\left(\beta-\frac{1}{\alpha}\right)$ is
$x^{2}-\left\{\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)\right\} x$
$+\left\{\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)\right\}=0$
On putting the values from Eqs. (i) and (ii), we get
$x^{2}-\frac{a}{b}(1-b) x+\frac{1}{b}(b-1)^{2}=0$
$\Rightarrow \quad b x^{2}+a(b-1) x+(b-1)^{2}=0, b \neq 0$