# Find whether the following equations have real roots. If real roots exist, find them.$\frac{1}{2 x-3}+\frac{1}{x-5}=1, x â‰ \frac{3}{2}, 5$

Given:

Given quadratic equation is $\frac{1}{2 x-3}+\frac{1}{x-5}=1, x ≠ \frac{3}{2}, 5$.

To do:

We have to determine whether the given quadratic equation has real roots.

Solution:

$\frac{1}{2 x-3}+\frac{1}{x-5}=1$

$\frac{x-5+2 x-3}{(2 x-3)(x-5)}=1$

$\frac{3 x-8}{2 x^{2}-3x-10 x+15}=1$

$\frac{3 x-8}{2 x^{2}-13x+15}=1$

$3 x-8 =2 x^{2}-13 x+15$

$2 x^{2}-13 x-3 x+15+8 =0$

$2 x^{2}-16x+23=0$

Comparing the above quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=2, b=-16$ and $c=23$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is

$D=b^2-4ac$.

Therefore,

$D=(-16)^2-4(2)(23)$

$=256-184$

$=72$

As $D>0$, the given quadratic equation has two distinct real roots.

This implies,

$x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-(-16) \pm \sqrt{72}}{2(2)}$

$x=\frac{16 \pm 6\sqrt{2}}{4}$

$x=\frac{2(8+3\sqrt{2})}{4}$ or $x= \frac{2(8-3\sqrt{2})}{4}$

$x=\frac{8+3\sqrt{2}}{2}$ or $x=\frac{8-3\sqrt{2}}{2}$

The roots of the given quadratic equation are $\frac{8+3\sqrt{2}}{2}$ and $\frac{8-3\sqrt{2}}{2}$.â€Šâ€Š

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