Solve 2=(1-2q)^2-2q^2

$$2 = ( 1 - 2 q ) ^ { 2 } - 2 q ^ { 2 }$$

Solve for q

$$q=\frac{\sqrt{6}}{2}+1\approx 2.224744871$$

$$q=-\frac{\sqrt{6}}{2}+1\approx -0.224744871$$

Steps Using the Quadratic Formula

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

$$2=1-4q+4q^{2}-2q^{2}$$

Combine 4q^{2} and -2q^{2} to get 2q^{2}.

$$2=1-4q+2q^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$1-4q+2q^{2}=2$$

Subtract 2 from both sides.

$$1-4q+2q^{2}-2=0$$

Subtract 2 from 1 to get -1.

$$-1-4q+2q^{2}=0$$

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.

$$2q^{2}-4q-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$q=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-1\right)}}{2\times 2}$$

Multiply -4 times 2.

$$q=\frac{-\left(-4\right)±\sqrt{16-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$q=\frac{-\left(-4\right)±\sqrt{16+8}}{2\times 2}$$

Add 16 to 8.

$$q=\frac{-\left(-4\right)±\sqrt{24}}{2\times 2}$$

Take the square root of 24.

$$q=\frac{-\left(-4\right)±2\sqrt{6}}{2\times 2}$$

The opposite of -4 is 4.

$$q=\frac{4±2\sqrt{6}}{2\times 2}$$

Multiply 2 times 2.

$$q=\frac{4±2\sqrt{6}}{4}$$

Now solve the equation q=\frac{4±2\sqrt{6}}{4} when ± is plus. Add 4 to 2\sqrt{6}.

$$q=\frac{2\sqrt{6}+4}{4}$$

Divide 4+2\sqrt{6} by 4.

$$q=\frac{\sqrt{6}}{2}+1$$

Now solve the equation q=\frac{4±2\sqrt{6}}{4} when ± is minus. Subtract 2\sqrt{6} from 4.

$$q=\frac{4-2\sqrt{6}}{4}$$

Divide 4-2\sqrt{6} by 4.

$$q=-\frac{\sqrt{6}}{2}+1$$

The equation is now solved.

$$q=\frac{\sqrt{6}}{2}+1 q=-\frac{\sqrt{6}}{2}+1$$

Steps for Completing the Square

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

$$2=1-4q+4q^{2}-2q^{2}$$

Combine 4q^{2} and -2q^{2} to get 2q^{2}.

$$2=1-4q+2q^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$1-4q+2q^{2}=2$$

Subtract 1 from both sides.

$$-4q+2q^{2}=2-1$$

Subtract 1 from 2 to get 1.

$$-4q+2q^{2}=1$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2q^{2}-4q=1$$

Divide both sides by 2.

$$\frac{2q^{2}-4q}{2}=\frac{1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$q^{2}+\left(-\frac{4}{2}\right)q=\frac{1}{2}$$

Divide -4 by 2.

$$q^{2}-2q=\frac{1}{2}$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}-2q+1=\frac{1}{2}+1$$

Add \frac{1}{2} to 1.

$$q^{2}-2q+1=\frac{3}{2}$$

Factor q^{2}-2q+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q-1\right)^{2}=\frac{3}{2}$$

Take the square root of both sides of the equation.

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

Simplify.

$$q-1=\frac{\sqrt{6}}{2} q-1=-\frac{\sqrt{6}}{2}$$

Add 1 to both sides of the equation.

$$q=\frac{\sqrt{6}}{2}+1 q=-\frac{\sqrt{6}}{2}+1$$