Question

Solve for x : (x^2+2)^2 + 8x^2 = 6x(x^2+2)
The answers given are 1 ± i , 2 ± √2

Answer 1

\[\frac{-[-(3+\sqrt{17})]\pm \sqrt{(3+\sqrt{17})^2-4(1)(2)} }{ 2(1) }\]

Answer 2

\[\frac{-[-(3+\sqrt{17})]\pm \sqrt{9+6\sqrt{17}+17-8} }{ 2 }\]

Answer 3

\[\frac{3+\sqrt{17}\pm \sqrt{6(3+\sqrt{17})} }{ 2 }\]

Answer 4

@ksananthu : is this what you meant ?

Answer 5

i'm sorry there is a mistake in my calculation

Answer 6

wait i will correct now

Answer 7

(x^2 + 2)^2 - 2*3 x (x^2 + 2) + 8 x^2 = 0
{(x^2 + 2)^2 - 2*3 x (x^2 + 2) + 9 x^2} +8 x^2 - 9 x^2 = 0
{(x^2 + 2) - 3 x}^2 - x^2 = 0
{(x^2 + 2) - 3 x}^2=x^2 taking roots bothsides
(x^2 + 2) - 3 x = x
x^2 - 4 x+2=0
\[x=\frac{ 4\pm \sqrt{16-4*1*2} }{ 2 }=2\pm \sqrt{2}\]

Answer 8

or
by taking roots we get another equation also
(x^2 + 2) - 3 x =- x
x^2 - 2 x+2=0
then it will give the value of x=1±i

Answer 9

now you got the answer

Answer 10

thanks :)