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The roots of the equation (x^2+1)^2=x(3x^2+4x+3) are given by
Dear Sudha,Are you sure you posted the right coefficients? Generally such questions with order greater than 2 have either integral solutions like -2,-1,1,2 etc or quick hint to factorization. In this question that is not happening. There is a root between 0 and 1 and one root between 3 and 4. Other 2 roots are imaginary. May be solvable but not beautiful in any way..
Please feel free to post as many doubts on our discussion forum as you can.we will get you the answer and detailed solution very quickly. All the best.Regards,Askiitians Experts,Gokul Joshi
Thanks
Hello sudha
(x^2+1)^2 = x(3x^2 + 4x + 3)
(x^2+1)^2 = x(3x^2 +4x) + 3x
(x^2+1)^2 = 3x^3 + 4x^2 + 3x
(x^2+1)^2 = 3x(x^2 + 1) + 4x^2
(x^2 + 1)^2 = 3x.(x^2+1) -9/4x^2 + 25x^2/4
(x^2+1)^2 - 3x.(x^2+1) + 9x^2/4 = 25x^2/4
( x ^2 + 1 - 3x/2 ) ^ 2 = 25x^2/4
So x^2 -3x/2 + 1 = + 5x/2 and x^2 -3x/2 + 1 = -5x/2
x^2 -4x + 1 = 0 and x^2 + x + 1 = 0
x^2 -4x + 4 = 3
(x-2)^2 = 3
or x = 2 +root(3) and x = 2 +root(3) Here x is unreal root ...with x = w and w^2 (w is cube root od unity)
If you get my answer ...please comment ..
With regards
Yagya
askiitians_expert
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