Question

x+1/x-1 + x-2/x+2 =4-2x+3/x-2, find x​

Answer 1

Answer:

Step-by-step explanation: Givenx−2x+1+x+2x−2=4−x−22x+3

\begin{lgathered}\implies \frac{x-2}{x+2}+\frac{2x+3}{x-2}\\=4-\frac{x+1}{x-1}\end{lgathered}⟹x+2x−2+x−22x+3=4−x−1x+1

\begin{lgathered}\implies \frac{(x-2)^{2}+(x+2)(2x+3)}{(x+2)(x-2)}\\=\frac{4(x-1)-(x+1)}{x-1}\end{lgathered}⟹(x+2)(x−2)(x−2)2+(x+2)(2x+3)=x−14(x−1)−(x+1)

\begin{lgathered}\implies \frac{x^{2}-4x+4+2x^{2}+3x+4x+6}{x^{2}-2^{2}}\\=\frac{4x-4-x-1}{x-1}\end{lgathered}⟹x2−22x2−4x+4+2x2+3x+4x+6=x−14x−4−x−1

\begin{lgathered}\implies \frac{3x^{2}+3x+10}{x^{2}-4}\\=\frac{3x-5}{x-1}\end{lgathered}⟹x2−43x2+3x+10=x−13x−5

\begin{lgathered}\implies (x-1)(3x^{2}+3x+10)\\=(x^{2}-4)(3x-5)\end{lgathered}⟹(x−1)(3x2+3x+10)=(x2−4)(3x−5)

\begin{lgathered}\implies (3x^{3}+3x^{2}+10x-3x^{2}-3x-10)\\=(3x^{3}-5x^{2}-12x+20)\end{lgathered}⟹(3x3+3x2+10x−3x2−3x−10)=(3x3−5x2−12x+20)

\implies (3x^{3}+7x-10)-(3x^{3}-5x^{2}-12x+20)=0⟹(3x3+7x−10)−(3x3−5x2−12x+20)=0

\implies (3x^{3}+7x-10-3x^{3}+5x^{2}+12x-20)=0⟹(3x3+7x−10−3x3+5x2+12x−20)=0

\implies 5x^{2}+19x-30=0⟹5x2+19x−30=0

Compare the 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}2×5−19±961

=\frac{-19±31}{10}10−19±31

Now,

x= \frac{-19+31}{10}10−19+31

Or x =\frac{-19-31}{10}10−19−31

x = \frac{12}{10}\: Or\: x =\frac{-50}{10}1012Orx=10−50

x = \frac{6}{5}\: Or \:x =-556Orx=−5

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