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Find the roots of the following equations:
(i) $ x-\frac{1}{x}=3, x ≠0 $
(ii) $ \frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x ≠-4,7 $
To do:
We have to find the roots of the given equations.
Solution:
(i) \( x-\frac{1}{x}=3, x ≠ 0 \)
$\frac{x}{1}-\frac{1}{x}=3$
$\frac{x^{2}-1}{x}=3$
$x^{2}-1=3 x$
$x^{2}-3 x-1=0$
The above equation is of the form $a x^{2}+b x+c=0$, where $a=1, b=-3$ and $c=-1$
Discriminant $\mathrm{D}=b^{2}-4 a c$
$=(-3)^{2}-4 \times 1 \times(-1)$
$=9+4$
$=13$
Let the roots of the above quadratic equation be $\alpha$ and $\beta$
Therefore,
$\alpha=\frac{-b+\sqrt{\mathrm{D}}}{2 a}$
$=\frac{-(-3)+\sqrt{13}}{2 \times 1}$
$=\frac{3+\sqrt{13}}{2}$
$\beta=\frac{-b-\sqrt{\mathrm{D}}}{2 a}$
$=\frac{-(-3)-\sqrt{13}}{2 \times 1}$
$=\frac{3-\sqrt{13}}{2}$
Hence, the roots of the given equation are $\frac{3+\sqrt{13}}{2}, \frac{3-\sqrt{13}}{2}$
(ii) \( \frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x ≠ -4,7 \)
$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$
$\frac{x-7-(x+4)}{(x+4)(x-7)}=\frac{11}{30}$
$\frac{x-7-x-4}{x^{2}-7 x+4 x-28}=\frac{11}{30}$
$\frac{-11}{x^{2}-3 x-28}=\frac{11}{30}$
$\frac{-1}{x^{2}-3 x-28}=\frac{1}{30}$
$x^{2}-3 x-28+30=0$
$x^{2}-3 x+2=0$
The above equation is of the form $a x^{2}+b x+c=0$, where $a=1, b=-3$ and $c=2$
Discriminant $D =b^{2}-4 a c$
$=(-3)^{2}-4 \times 1 \times 2$
$=9-8$
$=1$
Let the roots of the above quadratic equation be $\alpha$ and $\beta$
Therefore,
$\alpha=\frac{-b+\sqrt{D}}{2 a}$
$=\frac{-(-3)+\sqrt{1}}{2 \times 1}$
$=\frac{3+1}{2}$
$=\frac{4}{2}$
$=2$
$\beta=\frac{-b-\sqrt{D}}{2 a}$
$=\frac{-(-3)-\sqrt{1}}{2 \times 1}$
$=\frac{3-1}{2}$
$=\frac{2}{2}$
$=1$
Hence, the roots of the given equation are $1$ and $2$.