Question

Find the roots of x^4+x^2-12=0

Answer 1

`x^4+x^2-12=0`

Let:

y = x²

so:

`y^2+y-12=0`

Factor:

The factors of -12 that sum to 1 are -3 and 4, therefore:

`\begin{gathered} (y-3)(y+4)=0 \\ \end{gathered}`

Split into 2 equations:

`\begin{gathered} y-3=0 \\ and \\ y+4=0 \\ _{\text{ }}where\colon \\ y=x^2 \end{gathered}`

Therefore:

`\begin{gathered} x^2=3 \\ x=\pm\sqrt[]{3} \\ ---- \\ x^2=-4 \end{gathered}`

x² = 4 has no solution since for all x on the real line:

`\begin{gathered} x^2\ge0 \\ \end{gathered}`

Therefore, the roots are:

`\begin{gathered} x=\sqrt[]{3}\approx1.732 \\ or \\ x=-\sqrt[]{3}\approx-1.732 \end{gathered}`