Question

Using theproperties of proportion solve of x, given
$\frac{ {x}^{4} + 1 }{ {2x}^{2} } = \frac{17}{8}$

Answer 1

8[x⁴+1]=17×2x²

first solution

8x⁴+8=34x²

8x⁴-34x²+8=0

4x⁴-17x²+4=0

4x⁴-16x²-x²+4=0

4x²(x²-4)-1(x²-4)=0

(4x²-1)(x²-4)=0
[((2x)²-1²)(x²-2²)]=0

(2x-1)(2x+1)(x-2)(x+2)=0

x= -1/2,+1/2,-2,+2