Solve (3/2x-3)^2=2x

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

Solve for x

$$x = \frac{4 \sqrt{10} + 22}{9} \approx 3.849901182$$

$$x = \frac{22 - 4 \sqrt{10}}{9} \approx 1.038987707$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{2}x-3\right)^{2}.

$$\frac{9}{4}x^{2}-9x+9=2x$$

Subtract 2x from both sides.

$$\frac{9}{4}x^{2}-9x+9-2x=0$$

Combine -9x and -2x to get -11x.

$$\frac{9}{4}x^{2}-11x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{9}{4} for a, -11 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times \frac{9}{4}\times 9}}{2\times \frac{9}{4}}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times \frac{9}{4}\times 9}}{2\times \frac{9}{4}}$$

Multiply -4 times \frac{9}{4}.

$$x=\frac{-\left(-11\right)±\sqrt{121-9\times 9}}{2\times \frac{9}{4}}$$

Multiply -9 times 9.

$$x=\frac{-\left(-11\right)±\sqrt{121-81}}{2\times \frac{9}{4}}$$

Add 121 to -81.

$$x=\frac{-\left(-11\right)±\sqrt{40}}{2\times \frac{9}{4}}$$

Take the square root of 40.

$$x=\frac{-\left(-11\right)±2\sqrt{10}}{2\times \frac{9}{4}}$$

The opposite of -11 is 11.

$$x=\frac{11±2\sqrt{10}}{2\times \frac{9}{4}}$$

Multiply 2 times \frac{9}{4}.

$$x=\frac{11±2\sqrt{10}}{\frac{9}{2}}$$

Now solve the equation x=\frac{11±2\sqrt{10}}{\frac{9}{2}} when ± is plus. Add 11 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}+11}{\frac{9}{2}}$$

Divide 11+2\sqrt{10} by \frac{9}{2} by multiplying 11+2\sqrt{10} by the reciprocal of \frac{9}{2}.

$$x=\frac{4\sqrt{10}+22}{9}$$

Now solve the equation x=\frac{11±2\sqrt{10}}{\frac{9}{2}} when ± is minus. Subtract 2\sqrt{10} from 11.

$$x=\frac{11-2\sqrt{10}}{\frac{9}{2}}$$

Divide 11-2\sqrt{10} by \frac{9}{2} by multiplying 11-2\sqrt{10} by the reciprocal of \frac{9}{2}.

$$x=\frac{22-4\sqrt{10}}{9}$$

The equation is now solved.

$$x=\frac{4\sqrt{10}+22}{9} x=\frac{22-4\sqrt{10}}{9}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{2}x-3\right)^{2}.

$$\frac{9}{4}x^{2}-9x+9=2x$$

Subtract 2x from both sides.

$$\frac{9}{4}x^{2}-9x+9-2x=0$$

Combine -9x and -2x to get -11x.

$$\frac{9}{4}x^{2}-11x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$\frac{9}{4}x^{2}-11x=-9$$

Divide both sides of the equation by \frac{9}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{9}{4}x^{2}-11x}{\frac{9}{4}}=-\frac{9}{\frac{9}{4}}$$

Dividing by \frac{9}{4} undoes the multiplication by \frac{9}{4}.

$$x^{2}+\left(-\frac{11}{\frac{9}{4}}\right)x=-\frac{9}{\frac{9}{4}}$$

Divide -11 by \frac{9}{4} by multiplying -11 by the reciprocal of \frac{9}{4}.

$$x^{2}-\frac{44}{9}x=-\frac{9}{\frac{9}{4}}$$

Divide -9 by \frac{9}{4} by multiplying -9 by the reciprocal of \frac{9}{4}.

$$x^{2}-\frac{44}{9}x=-4$$

Divide -\frac{44}{9}, the coefficient of the x term, by 2 to get -\frac{22}{9}. Then add the square of -\frac{22}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{44}{9}x+\left(-\frac{22}{9}\right)^{2}=-4+\left(-\frac{22}{9}\right)^{2}$$

Square -\frac{22}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{44}{9}x+\frac{484}{81}=-4+\frac{484}{81}$$

Add -4 to \frac{484}{81}.

$$x^{2}-\frac{44}{9}x+\frac{484}{81}=\frac{160}{81}$$

Factor x^{2}-\frac{44}{9}x+\frac{484}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{22}{9}\right)^{2}=\frac{160}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{22}{9}\right)^{2}}=\sqrt{\frac{160}{81}}$$

Simplify.

$$x-\frac{22}{9}=\frac{4\sqrt{10}}{9} x-\frac{22}{9}=-\frac{4\sqrt{10}}{9}$$

Add \frac{22}{9} to both sides of the equation.

$$x=\frac{4\sqrt{10}+22}{9} x=\frac{22-4\sqrt{10}}{9}$$