Solve sqrt{x^4+9}-sqrt{6x^2+1}=0

$$\sqrt { x ^ { 4 } + 9 } - \sqrt { 6 x ^ { 2 } + 1 } = 0$$

Solve for x

$$x=-\sqrt{2}\approx -1.414213562$$

$$x=\sqrt{2}\approx 1.414213562$$

$$x=2$$

$$x=-2$$

Solution Steps

Subtract -\sqrt{6x^{2}+1} from both sides of the equation.

$$\sqrt{x^{4}+9}=\sqrt{6x^{2}+1}$$

Square both sides of the equation.

$$\left(\sqrt{x^{4}+9}\right)^{2}=\left(\sqrt{6x^{2}+1}\right)^{2}$$

Calculate \sqrt{x^{4}+9} to the power of 2 and get x^{4}+9.

$$x^{4}+9=\left(\sqrt{6x^{2}+1}\right)^{2}$$

Calculate \sqrt{6x^{2}+1} to the power of 2 and get 6x^{2}+1.

$$x^{4}+9=6x^{2}+1$$

Subtract 6x^{2} from both sides.

$$x^{4}+9-6x^{2}=1$$

Subtract 1 from both sides.

$$x^{4}+9-6x^{2}-1=0$$

Subtract 1 from 9 to get 8.

$$x^{4}+8-6x^{2}=0$$

Substitute t for x^{2}.

$$t^{2}-6t+8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -6 for b, and 8 for c in the quadratic formula.

$$t=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 8}}{2}$$

Do the calculations.

$$t=\frac{6±2}{2}$$

Solve the equation t=\frac{6±2}{2} when ± is plus and when ± is minus.

$$t=4 t=2$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for each t.

$$x=2 x=-2 x=\sqrt{2} x=-\sqrt{2}$$

Substitute 2 for x in the equation \sqrt{x^{4}+9}-\sqrt{6x^{2}+1}=0.

$$\sqrt{2^{4}+9}-\sqrt{6\times 2^{2}+1}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

Substitute -2 for x in the equation \sqrt{x^{4}+9}-\sqrt{6x^{2}+1}=0.

$$\sqrt{\left(-2\right)^{4}+9}-\sqrt{6\left(-2\right)^{2}+1}=0$$

Simplify. The value x=-2 satisfies the equation.

$$0=0$$

Substitute \sqrt{2} for x in the equation \sqrt{x^{4}+9}-\sqrt{6x^{2}+1}=0.

$$\sqrt{\left(\sqrt{2}\right)^{4}+9}-\sqrt{6\left(\sqrt{2}\right)^{2}+1}=0$$

Simplify. The value x=\sqrt{2} satisfies the equation.

$$0=0$$

Substitute -\sqrt{2} for x in the equation \sqrt{x^{4}+9}-\sqrt{6x^{2}+1}=0.

$$\sqrt{\left(-\sqrt{2}\right)^{4}+9}-\sqrt{6\left(-\sqrt{2}\right)^{2}+1}=0$$

Simplify. The value x=-\sqrt{2} satisfies the equation.

$$0=0$$

List all solutions of \sqrt{x^{4}+9}=\sqrt{6x^{2}+1}.

$$x=2 x=-2 x=\sqrt{2} x=-\sqrt{2}$$