Solve (2x+1/2)^2=4x

$$( 2 x + \frac { 1 } { 2 } ) ^ { 2 } = 4 x$$

Solve for x

$$x=\frac{1}{4}=0.25$$

Steps Using the Quadratic Formula

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

$$4x^{2}+2x+\frac{1}{4}=4x$$

Subtract 4x from both sides.

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

Combine 2x and -4x to get -2x.

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

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

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

Square -2.

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

Multiply -4 times 4.

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

Multiply -16 times \frac{1}{4}.

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

Add 4 to -4.

$$x=\frac{-\left(-2\right)±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$x=-\frac{-2}{2\times 4}$$

The opposite of -2 is 2.

$$x=\frac{2}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{2}{8}$$

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

$$x=\frac{1}{4}$$

Steps for Completing the Square

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

$$4x^{2}+2x+\frac{1}{4}=4x$$

Subtract 4x from both sides.

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

Combine 2x and -4x to get -2x.

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

Subtract \frac{1}{4} from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{1}{4} by 4.

$$x^{2}-\frac{1}{2}x=-\frac{1}{16}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-\frac{1}{16}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{-1+1}{16}$$

Add -\frac{1}{16} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=0$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{1}{4}=0 x-\frac{1}{4}=0$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{1}{4} x=\frac{1}{4}$$

The equation is now solved. Solutions are the same.

$$x=\frac{1}{4}$$