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]

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All the best Vignesh!!!



Regards,

Askiitians Experts

Priyansh Bajaj

Approved

     =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