Solve 2sqrt{x^2-4}=4-x

$$2 \sqrt { x ^ { 2 } - 4 } = 4 - x$$

Solve for x

$$x=\frac{-4\sqrt{7}-4}{3}\approx -4.861001748$$

$$x = \frac{4 \sqrt{7} - 4}{3} \approx 2.194335081$$

Solution Steps

Square both sides of the equation.

$$\left(2\sqrt{x^{2}-4}\right)^{2}=\left(4-x\right)^{2}$$

Expand \left(2\sqrt{x^{2}-4}\right)^{2}.

$$2^{2}\left(\sqrt{x^{2}-4}\right)^{2}=\left(4-x\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x^{2}-4}\right)^{2}=\left(4-x\right)^{2}$$

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

$$4\left(x^{2}-4\right)=\left(4-x\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}-4.

$$4x^{2}-16=\left(4-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$4x^{2}-16=16-8x+x^{2}$$

Subtract 16 from both sides.

$$4x^{2}-16-16=-8x+x^{2}$$

Subtract 16 from -16 to get -32.

$$4x^{2}-32=-8x+x^{2}$$

Add 8x to both sides.

$$4x^{2}-32+8x=x^{2}$$

Subtract x^{2} from both sides.

$$4x^{2}-32+8x-x^{2}=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-32+8x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$3x^{2}+8x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 8 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\left(-32\right)}}{2\times 3}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 3\left(-32\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-8±\sqrt{64-12\left(-32\right)}}{2\times 3}$$

Multiply -12 times -32.

$$x=\frac{-8±\sqrt{64+384}}{2\times 3}$$

Add 64 to 384.

$$x=\frac{-8±\sqrt{448}}{2\times 3}$$

Take the square root of 448.

$$x=\frac{-8±8\sqrt{7}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-8±8\sqrt{7}}{6}$$

Now solve the equation x=\frac{-8±8\sqrt{7}}{6} when ± is plus. Add -8 to 8\sqrt{7}.

$$x=\frac{8\sqrt{7}-8}{6}$$

Divide -8+8\sqrt{7} by 6.

$$x=\frac{4\sqrt{7}-4}{3}$$

Now solve the equation x=\frac{-8±8\sqrt{7}}{6} when ± is minus. Subtract 8\sqrt{7} from -8.

$$x=\frac{-8\sqrt{7}-8}{6}$$

Divide -8-8\sqrt{7} by 6.

$$x=\frac{-4\sqrt{7}-4}{3}$$

The equation is now solved.

$$x=\frac{4\sqrt{7}-4}{3} x=\frac{-4\sqrt{7}-4}{3}$$

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

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

Simplify. The value x=\frac{4\sqrt{7}-4}{3} satisfies the equation.

$$\frac{16}{3}-\frac{4}{3}\times 7^{\frac{1}{2}}=\frac{16}{3}-\frac{4}{3}\times 7^{\frac{1}{2}}$$

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

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

Simplify. The value x=\frac{-4\sqrt{7}-4}{3} satisfies the equation.

$$\frac{16}{3}+\frac{4}{3}\times 7^{\frac{1}{2}}=\frac{16}{3}+\frac{4}{3}\times 7^{\frac{1}{2}}$$

List all solutions of 2\sqrt{x^{2}-4}=4-x.

$$x=\frac{4\sqrt{7}-4}{3} x=\frac{-4\sqrt{7}-4}{3}$$