Solve the following quadratic equation by factorization:
$\frac{x\ +\ 3}{x\ -\ 2}\ -\ \frac{1\ -\ x}{x}\ =\ \frac{17}{4},\ x\ ≠\ 0,\ 2$

Given:

Given quadratic equation is $\frac{x\ +\ 3}{x\ -\ 2}\ -\ \frac{1\ -\ x}{x}\ =\ \frac{17}{4},\ x\ ≠\ 0,\ 2$.

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

$\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}$

$\frac{x(x+3)-(1-x)(x-2)}{(x-2)(x)}=\frac{17}{4}$

$\frac{x^2+3x-(x-2-x^2+2x)}{x^2-2x}=\frac{17}{4}$

$\frac{x^2+3x-3x+x^2+2}{x^2-2x}=\frac{17}{4}$

$4(2x^2+2)=17(x^2-2x)$   (On cross multiplication)

$8x^2+8=17x^2-34x$

$(17-8)x^2-34x-8=0$

$9x^2-34x-8=0$

$9x^2-36x+2x-8=0$   

$9x(x-4)+2(x-4)=0$

$(9x+2)(x-4)=0$

$9x+2=0$ or $x-4=0$

$9x=-2$ or $x=4$

$x=-\frac{2}{9}$ or $x=4$

The roots of the given quadratic equation are $-\frac{2}{9}$ and $4$.

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Updated on: 10-Oct-2022

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