# How to solve a double quadratic inequality

I have a double quadratic inequality of the form $2x^4 + 2x^2 - 3 < 0$. Had it been equation rather than inequality I would have probably set $x^2 = z$ and looked for the solution in terms of z and finally in terms of x. But how to deal with this inequality?

You can complete the square: $$2\left(x^2+\frac{1}{2}\right)^2-\frac{7}{2}<0,$$ or $$\left(x^2+\frac{1}{2}\right)^2<\frac{7}{4}$$ Can you see how to solve this?
• Solving $(x^2+\frac{1}{2})^2=\frac{7}{4}$ gives $x^2=-\frac{1}{2}\pm\frac{\sqrt{7}}{2}$, so $-\frac{1}{2}-\frac{\sqrt{7}}{2}<x^2<-\frac{1}{2}+\frac{\sqrt{7}}{2}$. – A. Goodier Jan 7 '18 at 20:57
• You can ignore the lower bound, since $x^2\geq 0$ for all real numbers $x$. So you just need to solve $0\leq x^2<-\frac{1}{2}+\frac{\sqrt{7}}{2}$. – A. Goodier Jan 7 '18 at 20:58
It's $$x^2<\frac{-1+\sqrt{7}}{2}$$ or $$-\sqrt{\frac{-1+\sqrt{7}}{2}}<x<\sqrt{\frac{-1+\sqrt{7}}{2}}.$$