Solve (x-sqrt{5})(x+sqrt{5})+(2x-1)^2=0

$$( x - \sqrt { 5 } ) ( x + \sqrt { 5 } ) + ( 2 x - 1 ) ^ { 2 } = 0$$

Solve for x

$$x = \frac{2 \sqrt{6} + 2}{5} \approx 1.379795897$$

$$x=\frac{2-2\sqrt{6}}{5}\approx -0.579795897$$

Steps Using the Quadratic Formula

Consider \left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-\left(\sqrt{5}\right)^{2}+\left(2x-1\right)^{2}=0$$

The square of \sqrt{5} is 5.

$$x^{2}-5+\left(2x-1\right)^{2}=0$$

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

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

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

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

Add -5 and 1 to get -4.

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

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.

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

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 5\left(-4\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-4\right)±\sqrt{16-20\left(-4\right)}}{2\times 5}$$

Multiply -20 times -4.

$$x=\frac{-\left(-4\right)±\sqrt{16+80}}{2\times 5}$$

Add 16 to 80.

$$x=\frac{-\left(-4\right)±\sqrt{96}}{2\times 5}$$

Take the square root of 96.

$$x=\frac{-\left(-4\right)±4\sqrt{6}}{2\times 5}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{6}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{4±4\sqrt{6}}{10}$$

Now solve the equation x=\frac{4±4\sqrt{6}}{10} when ± is plus. Add 4 to 4\sqrt{6}.

$$x=\frac{4\sqrt{6}+4}{10}$$

Divide 4+4\sqrt{6} by 10.

$$x=\frac{2\sqrt{6}+2}{5}$$

Now solve the equation x=\frac{4±4\sqrt{6}}{10} when ± is minus. Subtract 4\sqrt{6} from 4.

$$x=\frac{4-4\sqrt{6}}{10}$$

Divide 4-4\sqrt{6} by 10.

$$x=\frac{2-2\sqrt{6}}{5}$$

The equation is now solved.

$$x=\frac{2\sqrt{6}+2}{5} x=\frac{2-2\sqrt{6}}{5}$$

Steps for Completing the Square

Consider \left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-\left(\sqrt{5}\right)^{2}+\left(2x-1\right)^{2}=0$$

The square of \sqrt{5} is 5.

$$x^{2}-5+\left(2x-1\right)^{2}=0$$

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

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

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

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

Add -5 and 1 to get -4.

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

Add 4 to both sides. Anything plus zero gives itself.

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

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

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.

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

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

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

Add \frac{4}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{24}{25}$$

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}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{24}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{24}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{2\sqrt{6}}{5} x-\frac{2}{5}=-\frac{2\sqrt{6}}{5}$$

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

$$x=\frac{2\sqrt{6}+2}{5} x=\frac{2-2\sqrt{6}}{5}$$