Solve x^2-2x+5/4=1/2x

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

Solve for x

$$x = \frac{\sqrt{5} + 5}{4} \approx 1.809016994$$

$$x=\frac{5-\sqrt{5}}{4}\approx 0.690983006$$

Steps Using the Quadratic Formula

Subtract \frac{1}{2}x from both sides.

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

Combine -2x and -\frac{1}{2}x to get -\frac{5}{2}x.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{5}{2} for b, and \frac{5}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

Multiply -4 times \frac{5}{4}.

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

Add \frac{25}{4} to -5.

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

Take the square root of \frac{5}{4}.

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

The opposite of -\frac{5}{2} is \frac{5}{2}.

$$x=\frac{\frac{5}{2}±\frac{\sqrt{5}}{2}}{2}$$

Now solve the equation x=\frac{\frac{5}{2}±\frac{\sqrt{5}}{2}}{2} when ± is plus. Add \frac{5}{2} to \frac{\sqrt{5}}{2}.

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

Divide \frac{5+\sqrt{5}}{2} by 2.

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

Now solve the equation x=\frac{\frac{5}{2}±\frac{\sqrt{5}}{2}}{2} when ± is minus. Subtract \frac{\sqrt{5}}{2} from \frac{5}{2}.

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

Divide \frac{5-\sqrt{5}}{2} by 2.

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

The equation is now solved.

$$x=\frac{\sqrt{5}+5}{4} x=\frac{5-\sqrt{5}}{4}$$

Steps for Completing the Square

Subtract \frac{1}{2}x from both sides.

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

Combine -2x and -\frac{1}{2}x to get -\frac{5}{2}x.

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

Subtract \frac{5}{4} from both sides. Anything subtracted from zero gives its negation.

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

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

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

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{5}{4}+\frac{25}{16}$$

Add -\frac{5}{4} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{5}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{\sqrt{5}+5}{4} x=\frac{5-\sqrt{5}}{4}$$