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Find the square root of 1+ ((8x^3 + 8x) / (x^4 - 4x^3 + 5x^2 - 4x +4))
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]
Please feel free to post as many doubts on our discussion forum as you can. If you find any questionDifficult 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.All the best Vignesh!!!Regards,Askiitians ExpertsPriyansh Bajaj
=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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