# 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 $2$.

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