Solve x^2+(2x-1)^2=1

$$x ^ { 2 } + ( 2 x - 1 ) ^ { 2 } = 1$$

Solve for x

$$x=\frac{4}{5}=0.8$$

$$x=0$$

Steps Using Factoring

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

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

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-4x+1=1$$

Subtract 1 from both sides.

$$5x^{2}-4x+1-1=0$$

Subtract 1 from 1 to get 0.

$$5x^{2}-4x=0$$

Factor out x.

$$x\left(5x-4\right)=0$$

To find equation solutions, solve x=0 and 5x-4=0.

$$x=0 x=\frac{4}{5}$$

Steps Using the Quadratic Formula

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

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

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-4x+1=1$$

Subtract 1 from both sides.

$$5x^{2}-4x+1-1=0$$

Subtract 1 from 1 to get 0.

$$5x^{2}-4x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\times 5}$$

Take the square root of \left(-4\right)^{2}.

$$x=\frac{-\left(-4\right)±4}{2\times 5}$$

The opposite of -4 is 4.

$$x=\frac{4±4}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{4±4}{10}$$

Now solve the equation x=\frac{4±4}{10} when ± is plus. Add 4 to 4.

$$x=\frac{8}{10}$$

Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{5}$$

Now solve the equation x=\frac{4±4}{10} when ± is minus. Subtract 4 from 4.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

The equation is now solved.

$$x=\frac{4}{5} x=0$$

Steps for Completing the Square

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

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

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-4x+1=1$$

Subtract 1 from both sides.

$$5x^{2}-4x=1-1$$

Subtract 1 from 1 to get 0.

$$5x^{2}-4x=0$$

Divide both sides by 5.

$$\frac{5x^{2}-4x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{4}{5}x=\frac{0}{5}$$

Divide 0 by 5.

$$x^{2}-\frac{4}{5}x=0$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=\left(-\frac{2}{5}\right)^{2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{4}{25}$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{2}{5} x-\frac{2}{5}=-\frac{2}{5}$$

Add \frac{2}{5} to both sides of the equation.

$$x=\frac{4}{5} x=0$$