Solve x^2-2/3+x/2=x

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

Solve for x

$$x = \frac{\sqrt{41} + 3}{4} \approx 2.350781059$$

$$x=\frac{3-\sqrt{41}}{4}\approx -0.850781059$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 2 is 6. Multiply \frac{x^{2}-2}{3} times \frac{2}{2}. Multiply \frac{x}{2} times \frac{3}{3}.

$$\frac{2\left(x^{2}-2\right)}{6}+\frac{3x}{6}=x$$

Since \frac{2\left(x^{2}-2\right)}{6} and \frac{3x}{6} have the same denominator, add them by adding their numerators.

$$\frac{2\left(x^{2}-2\right)+3x}{6}=x$$

Do the multiplications in 2\left(x^{2}-2\right)+3x.

$$\frac{2x^{2}-4+3x}{6}=x$$

Divide each term of 2x^{2}-4+3x by 6 to get \frac{1}{3}x^{2}-\frac{2}{3}+\frac{1}{2}x.

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

Subtract x from both sides.

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

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

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

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.

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

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

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

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

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

Multiply -4 times \frac{1}{3}.

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

Multiply -\frac{4}{3} times -\frac{2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}+\frac{8}{9}}}{2\times \frac{1}{3}}$$

Add \frac{1}{4} to \frac{8}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{41}{36}}}{2\times \frac{1}{3}}$$

Take the square root of \frac{41}{36}.

$$x=\frac{-\left(-\frac{1}{2}\right)±\frac{\sqrt{41}}{6}}{2\times \frac{1}{3}}$$

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

$$x=\frac{\frac{1}{2}±\frac{\sqrt{41}}{6}}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$x=\frac{\frac{1}{2}±\frac{\sqrt{41}}{6}}{\frac{2}{3}}$$

Now solve the equation x=\frac{\frac{1}{2}±\frac{\sqrt{41}}{6}}{\frac{2}{3}} when ± is plus. Add \frac{1}{2} to \frac{\sqrt{41}}{6}.

$$x=\frac{\frac{\sqrt{41}}{6}+\frac{1}{2}}{\frac{2}{3}}$$

Divide \frac{1}{2}+\frac{\sqrt{41}}{6} by \frac{2}{3} by multiplying \frac{1}{2}+\frac{\sqrt{41}}{6} by the reciprocal of \frac{2}{3}.

$$x=\frac{\sqrt{41}+3}{4}$$

Now solve the equation x=\frac{\frac{1}{2}±\frac{\sqrt{41}}{6}}{\frac{2}{3}} when ± is minus. Subtract \frac{\sqrt{41}}{6} from \frac{1}{2}.

$$x=\frac{-\frac{\sqrt{41}}{6}+\frac{1}{2}}{\frac{2}{3}}$$

Divide \frac{1}{2}-\frac{\sqrt{41}}{6} by \frac{2}{3} by multiplying \frac{1}{2}-\frac{\sqrt{41}}{6} by the reciprocal of \frac{2}{3}.

$$x=\frac{3-\sqrt{41}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{41}+3}{4} x=\frac{3-\sqrt{41}}{4}$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 2 is 6. Multiply \frac{x^{2}-2}{3} times \frac{2}{2}. Multiply \frac{x}{2} times \frac{3}{3}.

$$\frac{2\left(x^{2}-2\right)}{6}+\frac{3x}{6}=x$$

Since \frac{2\left(x^{2}-2\right)}{6} and \frac{3x}{6} have the same denominator, add them by adding their numerators.

$$\frac{2\left(x^{2}-2\right)+3x}{6}=x$$

Do the multiplications in 2\left(x^{2}-2\right)+3x.

$$\frac{2x^{2}-4+3x}{6}=x$$

Divide each term of 2x^{2}-4+3x by 6 to get \frac{1}{3}x^{2}-\frac{2}{3}+\frac{1}{2}x.

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

Subtract x from both sides.

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

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

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

Add \frac{2}{3} to both sides. Anything plus zero gives itself.

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

Multiply both sides by 3.

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

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

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

Divide -\frac{1}{2} by \frac{1}{3} by multiplying -\frac{1}{2} by the reciprocal of \frac{1}{3}.

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

Divide \frac{2}{3} by \frac{1}{3} by multiplying \frac{2}{3} by the reciprocal of \frac{1}{3}.

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

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.

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

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

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

Add 2 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{41}{16}$$

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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{41}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{41}}{4} x-\frac{3}{4}=-\frac{\sqrt{41}}{4}$$

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

$$x=\frac{\sqrt{41}+3}{4} x=\frac{3-\sqrt{41}}{4}$$