Question

x+1/x-1 +x-2/x+2 =4(2x+3)/x-2
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Answer 1

$\begin{gathered} Given \: \frac{x+1}{x-2}+\frac{x-2}{x+2}\\=4-\frac{2x+3}{x-2}\end{gathered}$

$\begin{gathered} \: \frac{x+1}{x-2}+\frac{x-2}{x+2}\\=4-\frac{2x+3}{x-2}\end{gathered}$

$\begin{gathered}\implies \frac{x-2}{x+2}+\frac{2x+3}{x-2}\\=4-\frac{x+1}{x-1}\end{gathered}$

$\begin{gathered}\implies \frac{(x-2)^{2}+(x+2)(2x+3)}{(x+2)(x-2)}\\=\frac{4(x-1)-(x+1)}{x-1}\end{gathered}$

$\begin{gathered}\implies \frac{x^{2}-4x+4+2x^{2}+3x+4x+6}{x^{2}-2^{2}}\\=\frac{4x-4-x-1}{x-1}\end{gathered}$

$\begin{gathered}\implies \frac{3x^{2}+3x+10}{x^{2}-4}\\=\frac{3x-5}{x-1}\end{gathered}$

$\begin{gathered}\implies (x-1)(3x^{2}+3x+10)\\=(x^{2}-4)(3x-5)\end{gathered}$

$\begin{gathered}\implies (3x^{3}+3x^{2}+10x-3x^{2}-3x-10)\\=(3x^{3}-5x^{2}-12x+20)\end{gathered}$

$\implies (3x^{3}+7x-10)-(3x^{3}-5x^{2}-12x+20)=0$

$\implies (3x^{3}+7x-10-3x^{3}+5x^{2}+12x-20)=0$

$\implies 5x^{2}+19x-30=0$

Compare above equation with

ax²+bx+c=0, we get

a = 5, b = 19, c = -30

Discreminant (D)=b²-4ac

= (19)²-4×19×(-30)

= 361+600

= 961

By Quadratic Formula:

$x = [-b±√D]/(2a)$

$= \frac{-19±\sqrt{961}}{2\times 5}$

$=\frac{-19±31}{10}$

Now,

$x= \frac{-19+31}{10}$

$Or x =\frac{-19-31}{10}$

$x = \frac{12}{10}\: Or\: x =\frac{-50}{10}$

$x = \frac{6}{5}\: Or \:x =-5$