Solve x^2+3=1/2x

$$x ^ { 2 } + 3 = \frac { 1 } { 2 } x$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{47}i}{4}\approx 0.25+1.71391365i$$

$$x=\frac{-\sqrt{47}i+1}{4}\approx 0.25-1.71391365i$$

Steps Using the Quadratic Formula

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

$$x^{2}+3-\frac{1}{2}x=0$$

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.

$$x^{2}-\frac{1}{2}x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{1}{2} for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\left(-\frac{1}{2}\right)^{2}-4\times 3}}{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-4\times 3}}{2}$$

Multiply -4 times 3.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-12}}{2}$$

Add \frac{1}{4} to -12.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{-\frac{47}{4}}}{2}$$

Take the square root of -\frac{47}{4}.

$$x=\frac{-\left(-\frac{1}{2}\right)±\frac{\sqrt{47}i}{2}}{2}$$

The opposite of -\frac{1}{2} is \frac{1}{2}.

$$x=\frac{\frac{1}{2}±\frac{\sqrt{47}i}{2}}{2}$$

Now solve the equation x=\frac{\frac{1}{2}±\frac{\sqrt{47}i}{2}}{2} when ± is plus. Add \frac{1}{2} to \frac{i\sqrt{47}}{2}.

$$x=\frac{1+\sqrt{47}i}{2\times 2}$$

Divide \frac{1+i\sqrt{47}}{2} by 2.

$$x=\frac{1+\sqrt{47}i}{4}$$

Now solve the equation x=\frac{\frac{1}{2}±\frac{\sqrt{47}i}{2}}{2} when ± is minus. Subtract \frac{i\sqrt{47}}{2} from \frac{1}{2}.

$$x=\frac{-\sqrt{47}i+1}{2\times 2}$$

Divide \frac{1-i\sqrt{47}}{2} by 2.

$$x=\frac{-\sqrt{47}i+1}{4}$$

The equation is now solved.

$$x=\frac{1+\sqrt{47}i}{4} x=\frac{-\sqrt{47}i+1}{4}$$

Steps for Completing the Square

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

$$x^{2}+3-\frac{1}{2}x=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-\frac{1}{2}x=-3$$

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.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-3+\left(-\frac{1}{4}\right)^{2}$$

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-3+\frac{1}{16}$$

Add -3 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{47}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{47}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{47}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{47}i}{4} x-\frac{1}{4}=-\frac{\sqrt{47}i}{4}$$

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

$$x=\frac{1+\sqrt{47}i}{4} x=\frac{-\sqrt{47}i+1}{4}$$