Solve x^2-3sqrt{2x^2-7}=1

$$x ^ { 2 } - 3 \sqrt { 2 x ^ { 2 } - 7 } = 1$$

Solve for x

$$x=-4$$

$$x=4$$

$$x=2$$

$$x=-2$$

Solution Steps

Subtract x^{2} from both sides of the equation.

$$-3\sqrt{2x^{2}-7}=1-x^{2}$$

Square both sides of the equation.

$$\left(-3\sqrt{2x^{2}-7}\right)^{2}=\left(1-x^{2}\right)^{2}$$

Expand \left(-3\sqrt{2x^{2}-7}\right)^{2}.

$$\left(-3\right)^{2}\left(\sqrt{2x^{2}-7}\right)^{2}=\left(1-x^{2}\right)^{2}$$

Calculate -3 to the power of 2 and get 9.

$$9\left(\sqrt{2x^{2}-7}\right)^{2}=\left(1-x^{2}\right)^{2}$$

Calculate \sqrt{2x^{2}-7} to the power of 2 and get 2x^{2}-7.

$$9\left(2x^{2}-7\right)=\left(1-x^{2}\right)^{2}$$

Use the distributive property to multiply 9 by 2x^{2}-7.

$$18x^{2}-63=\left(1-x^{2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x^{2}\right)^{2}.

$$18x^{2}-63=1-2x^{2}+\left(x^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$18x^{2}-63=1-2x^{2}+x^{4}$$

Subtract 1 from both sides.

$$18x^{2}-63-1=-2x^{2}+x^{4}$$

Subtract 1 from -63 to get -64.

$$18x^{2}-64=-2x^{2}+x^{4}$$

Add 2x^{2} to both sides.

$$18x^{2}-64+2x^{2}=x^{4}$$

Combine 18x^{2} and 2x^{2} to get 20x^{2}.

$$20x^{2}-64=x^{4}$$

Subtract x^{4} from both sides.

$$20x^{2}-64-x^{4}=0$$

Substitute t for x^{2}.

$$-t^{2}+20t-64=0$$

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, 20 for b, and -64 for c in the quadratic formula.

$$t=\frac{-20±\sqrt{20^{2}-4\left(-1\right)\left(-64\right)}}{-2}$$

Do the calculations.

$$t=\frac{-20±12}{-2}$$

Solve the equation t=\frac{-20±12}{-2} when ± is plus and when ± is minus.

$$t=4 t=16$$

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

$$x=2 x=-2 x=4 x=-4$$

Substitute 2 for x in the equation x^{2}-3\sqrt{2x^{2}-7}=1.

$$2^{2}-3\sqrt{2\times 2^{2}-7}=1$$

Simplify. The value x=2 satisfies the equation.

$$1=1$$

Substitute -2 for x in the equation x^{2}-3\sqrt{2x^{2}-7}=1.

$$\left(-2\right)^{2}-3\sqrt{2\left(-2\right)^{2}-7}=1$$

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

$$1=1$$

Substitute 4 for x in the equation x^{2}-3\sqrt{2x^{2}-7}=1.

$$4^{2}-3\sqrt{2\times 4^{2}-7}=1$$

Simplify. The value x=4 satisfies the equation.

$$1=1$$

Substitute -4 for x in the equation x^{2}-3\sqrt{2x^{2}-7}=1.

$$\left(-4\right)^{2}-3\sqrt{2\left(-4\right)^{2}-7}=1$$

Simplify. The value x=-4 satisfies the equation.

$$1=1$$

List all solutions of -3\sqrt{2x^{2}-7}=1-x^{2}.

$$x=2 x=-2 x=4 x=-4$$