Solve 1/2x-3=x^2?

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

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 x^{2} from both sides.

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

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}.

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

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

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

Multiply -4 times -1.

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

Multiply 4 times -3.

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

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

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

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

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

Multiply 2 times -1.

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

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.

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

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.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Add 3 to both sides. Anything plus zero gives itself.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide \frac{1}{2} by -1.

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

Divide 3 by -1.

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

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.

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