Solve {l}{x^2=3/2x-1}{=}

$$\left. \begin{array} { l } { x ^ { 2 } = \frac { 3 } { 2 } x - 1 } \\ { = } \end{array} \right.$$

Solve for x

$$x=\frac{3+\sqrt{7}i}{4}\approx 0.75+0.661437828i$$

$$x=\frac{-\sqrt{7}i+3}{4}\approx 0.75-0.661437828i$$

Steps Using the Quadratic Formula

Subtract \frac{3}{2}x from both sides.

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

Add 1 to both sides.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-4}}{2}$$

Add \frac{9}{4} to -4.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{-\frac{7}{4}}}{2}$$

Take the square root of -\frac{7}{4}.

$$x=\frac{-\left(-\frac{3}{2}\right)±\frac{\sqrt{7}i}{2}}{2}$$

The opposite of -\frac{3}{2} is \frac{3}{2}.

$$x=\frac{\frac{3}{2}±\frac{\sqrt{7}i}{2}}{2}$$

Now solve the equation x=\frac{\frac{3}{2}±\frac{\sqrt{7}i}{2}}{2} when ± is plus. Add \frac{3}{2} to \frac{i\sqrt{7}}{2}.

$$x=\frac{3+\sqrt{7}i}{2\times 2}$$

Divide \frac{3+i\sqrt{7}}{2} by 2.

$$x=\frac{3+\sqrt{7}i}{4}$$

Now solve the equation x=\frac{\frac{3}{2}±\frac{\sqrt{7}i}{2}}{2} when ± is minus. Subtract \frac{i\sqrt{7}}{2} from \frac{3}{2}.

$$x=\frac{-\sqrt{7}i+3}{2\times 2}$$

Divide \frac{3-i\sqrt{7}}{2} by 2.

$$x=\frac{-\sqrt{7}i+3}{4}$$

The equation is now solved.

$$x=\frac{3+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+3}{4}$$

Steps for Completing the Square

Subtract \frac{3}{2}x from both sides.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-1+\frac{9}{16}$$

Add -1 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{7}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=-\frac{7}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{-\frac{7}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{7}i}{4} x-\frac{3}{4}=-\frac{\sqrt{7}i}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{3+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+3}{4}$$