Solve 2+2x=sqrt{6x^2+6}

$$2+2x= \sqrt{ 6 { x }^{ 2 } +6 }$$

Solve for x

$$x=\sqrt{3}+2\approx 3.732050808$$

$$x=2-\sqrt{3}\approx 0.267949192$$

Solution Steps

Square both sides of the equation.

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

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

$$4+8x+4x^{2}=\left(\sqrt{6x^{2}+6}\right)^{2}$$

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

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

Subtract 6x^{2} from both sides.

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

Combine 4x^{2} and -6x^{2} to get -2x^{2}.

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

Subtract 6 from both sides.

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

Subtract 6 from 4 to get -2.

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

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.

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

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

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

Square 8.

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

Multiply -4 times -2.

$$x=\frac{-8±\sqrt{64+8\left(-2\right)}}{2\left(-2\right)}$$

Multiply 8 times -2.

$$x=\frac{-8±\sqrt{64-16}}{2\left(-2\right)}$$

Add 64 to -16.

$$x=\frac{-8±\sqrt{48}}{2\left(-2\right)}$$

Take the square root of 48.

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

Multiply 2 times -2.

$$x=\frac{-8±4\sqrt{3}}{-4}$$

Now solve the equation x=\frac{-8±4\sqrt{3}}{-4} when ± is plus. Add -8 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}-8}{-4}$$

Divide -8+4\sqrt{3} by -4.

$$x=2-\sqrt{3}$$

Now solve the equation x=\frac{-8±4\sqrt{3}}{-4} when ± is minus. Subtract 4\sqrt{3} from -8.

$$x=\frac{-4\sqrt{3}-8}{-4}$$

Divide -8-4\sqrt{3} by -4.

$$x=\sqrt{3}+2$$

The equation is now solved.

$$x=2-\sqrt{3} x=\sqrt{3}+2$$

Substitute 2-\sqrt{3} for x in the equation 2+2x=\sqrt{6x^{2}+6}.

$$2+2\left(2-\sqrt{3}\right)=\sqrt{6\left(2-\sqrt{3}\right)^{2}+6}$$

Simplify. The value x=2-\sqrt{3} satisfies the equation.

$$6-2\times 3^{\frac{1}{2}}=6-2\times 3^{\frac{1}{2}}$$

Substitute \sqrt{3}+2 for x in the equation 2+2x=\sqrt{6x^{2}+6}.

$$2+2\left(\sqrt{3}+2\right)=\sqrt{6\left(\sqrt{3}+2\right)^{2}+6}$$

Simplify. The value x=\sqrt{3}+2 satisfies the equation.

$$6+2\times 3^{\frac{1}{2}}=6+2\times 3^{\frac{1}{2}}$$

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

$$x=2-\sqrt{3} x=\sqrt{3}+2$$