Solve x^2+x^2>9

$${ x }^{ 2 } + { x }^{ 2 } > 9$$

Solve for x

$$x\in \left(-\infty,-\frac{3\sqrt{2}}{2}\right)\cup \left(\frac{3\sqrt{2}}{2},\infty\right)$$

Solution Steps

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}>9$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x^{2}>\frac{9}{2}$$

Calculate the square root of \frac{9}{2} and get \frac{3\sqrt{2}}{2}. Rewrite \frac{9}{2} as \left(\frac{3\sqrt{2}}{2}\right)^{2}.

$$x^{2}>\left(\frac{3\sqrt{2}}{2}\right)^{2}$$

Inequality holds for |x|>\frac{3\sqrt{2}}{2}.

$$|x|>\frac{3\sqrt{2}}{2}$$

Rewrite |x|>\frac{3\sqrt{2}}{2} as x<-\frac{3\sqrt{2}}{2}\text{; }x>\frac{3\sqrt{2}}{2}.

$$x<-\frac{3\sqrt{2}}{2}\text{; }x>\frac{3\sqrt{2}}{2}$$