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`        Find the square root of 1+ ((8x^3 + 8x) / (x^4 - 4x^3 + 5x^2 - 4x +4))`
7 years ago

30 Points
```										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
```
7 years ago
akash kumar
16 Points
```										     =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

```
7 years ago
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