Find the real roots of \frac{18}{ {x}^(4) } + \frac{1}{ {x}^(2) } = 4 ​

Question

Find the real roots of

`\frac{18}{ {x}^(4) } + \frac{1}{ {x}^(2) } = 4`
​

Answer 1

see the photos above given

Answer 2

`$x^2=(9)/(4) \implies x=\pm (3)/(2) $`

`$x^2=-2 \implies x=\pm√(-2) \implies x=\pm√(2i) $`

The solutions are

`$\{-(3)/(2),(3)/(2), -√(2i), √(2i) \}$`

The Real roots are

`$\{-(3)/(2),(3)/(2)\}$`

Explanation:

`$(18)/(x^4) +(1)/(x^2)=4$`

Multiply both sides by
`x^4`

`$x^4(18)/(x^4) +x^4(1)/(x^2)=4\cdot x^4$`

`$18+x^2=4x^4$`

`4x^4-x^2-18=0`

Substitute
`x^2=t`

`4t^2-t-18=0`

Solving the quadratic equation using the quadratic formula:

`$t=(-(-1)\pm √((-1)^2-4\cdot 4(-18)))/(2\cdot 4)$`

`$t=(1\pm √(289))/(8)$`

`$t=(1\pm 17)/(8)$`

`$t_1=(9)/(4)$`

`t_2=-2`

Now we have to solve for
`x^2=t`

`$x^2=(9)/(4) \implies x=\pm (3)/(2) $`

`$x^2=-2 \implies x=\pm√(-2) \implies x=\pm√(2i) $`