Solve for $x$:
$\frac{1}{x}+\frac{2}{2x-3}=\frac{1}{x-2}, x≠0, \frac{3}{2}, 2$

Given:

Given quadratic equation is $\frac{1}{x}+\frac{2}{2x-3}=\frac{1}{x-2}, x≠0, \frac{3}{2}, 2$.

To do:

We have to solve the given quadratic equation.

Solution:

$\frac{1}{x}+\frac{2}{2x-3}=\frac{1}{x-2}, x≠0, \frac{3}{2}, 2$

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

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

$\frac{4x-3}{2x^2-3x}=\frac{1}{x-2}$

$(x-2)(4x-3)=1(2x^2-3x)$   (On cross multiplication)

$4x^2-3x-8x+6=2x^2-3x$

$(4-2)x^2+(-11+3)x+6=0$

$2x^2-8x+6=0$

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

$a=2, b=-8$ and $c=6$.

Therefore, the roots of the given equation are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-8)\pm \sqrt{(-8)^2-4(2)(6)}}{2(2)}$ 

$x=\frac{8\pm \sqrt{64-48}}{4}$ 

$x=\frac{8\pm \sqrt{16}}{4}$ 

$x=\frac{8\pm 4)}{4}$ 

$x=\frac{8+4}{4}$ or $x=\frac{8-4}{4}$

$x=\frac{12}{4}$ or $x=\frac{4}{4}$

$x=3$ or $x=1$

The values of $x$ are $1$ and $3$.

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Updated on: 10-Oct-2022

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