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# Find the square root of 1+ ((8x^3 + 8x) / (x^4 - 4x^3 + 5x^2 - 4x +4))

Grade:8

## 2 Answers

Priyansh Bajaj AskiitiansExpert-IITD
30 Points
11 years ago

Dear Vignesh,

Solution:- x4 -4x3 +5x2 -4x+4 = x4 + x2 -4x3 -4x +4x2 +4

= x2 (x2 +1) -4x(x2 +1) +4(x2 +1)

= (x2 +1)(x2 -4x +4)

= (x2 +1)(x -2)2

And, 8x3 + 8x = 8x(x2 +1)

To find sq. root of {1+ [8x(x2 +1)] / [(x2 +1)(x -2)2]}

or {1+ 8x/ (x -2)2}1/2 = {(x +2)2/ (x -2)2}1/2

= (x +2)/ (x -2) [ANS]

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Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.

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Regards,

Askiitians Experts

Priyansh Bajaj

akash kumar
16 Points
11 years ago

=1+(8x^3)/(x^4-4x^3+5x^2-4x+4

=(x^4-4x^3+5x^2-4x+4+8x^3+8x)/(x^4+4x^3+5x^2-4x+4

=-1

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