Solve (x-2)^2+(3/2x-3)^2=1

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

Solve for x

$$x=\frac{2\sqrt{13}}{13}+2\approx 2.554700196$$

$$x=-\frac{2\sqrt{13}}{13}+2\approx 1.445299804$$

Steps Using the Quadratic Formula

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

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

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

$$x^{2}-4x+4+\frac{9}{4}x^{2}-9x+9=1$$

Combine x^{2} and \frac{9}{4}x^{2} to get \frac{13}{4}x^{2}.

$$\frac{13}{4}x^{2}-4x+4-9x+9=1$$

Combine -4x and -9x to get -13x.

$$\frac{13}{4}x^{2}-13x+4+9=1$$

Add 4 and 9 to get 13.

$$\frac{13}{4}x^{2}-13x+13=1$$

Subtract 1 from both sides.

$$\frac{13}{4}x^{2}-13x+13-1=0$$

Subtract 1 from 13 to get 12.

$$\frac{13}{4}x^{2}-13x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{13}{4} for a, -13 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times \frac{13}{4}\times 12}}{2\times \frac{13}{4}}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times \frac{13}{4}\times 12}}{2\times \frac{13}{4}}$$

Multiply -4 times \frac{13}{4}.

$$x=\frac{-\left(-13\right)±\sqrt{169-13\times 12}}{2\times \frac{13}{4}}$$

Multiply -13 times 12.

$$x=\frac{-\left(-13\right)±\sqrt{169-156}}{2\times \frac{13}{4}}$$

Add 169 to -156.

$$x=\frac{-\left(-13\right)±\sqrt{13}}{2\times \frac{13}{4}}$$

The opposite of -13 is 13.

$$x=\frac{13±\sqrt{13}}{2\times \frac{13}{4}}$$

Multiply 2 times \frac{13}{4}.

$$x=\frac{13±\sqrt{13}}{\frac{13}{2}}$$

Now solve the equation x=\frac{13±\sqrt{13}}{\frac{13}{2}} when ± is plus. Add 13 to \sqrt{13}.

$$x=\frac{\sqrt{13}+13}{\frac{13}{2}}$$

Divide 13+\sqrt{13} by \frac{13}{2} by multiplying 13+\sqrt{13} by the reciprocal of \frac{13}{2}.

$$x=\frac{2\sqrt{13}}{13}+2$$

Now solve the equation x=\frac{13±\sqrt{13}}{\frac{13}{2}} when ± is minus. Subtract \sqrt{13} from 13.

$$x=\frac{13-\sqrt{13}}{\frac{13}{2}}$$

Divide 13-\sqrt{13} by \frac{13}{2} by multiplying 13-\sqrt{13} by the reciprocal of \frac{13}{2}.

$$x=-\frac{2\sqrt{13}}{13}+2$$

The equation is now solved.

$$x=\frac{2\sqrt{13}}{13}+2 x=-\frac{2\sqrt{13}}{13}+2$$

Steps for Completing the Square

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

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

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

$$x^{2}-4x+4+\frac{9}{4}x^{2}-9x+9=1$$

Combine x^{2} and \frac{9}{4}x^{2} to get \frac{13}{4}x^{2}.

$$\frac{13}{4}x^{2}-4x+4-9x+9=1$$

Combine -4x and -9x to get -13x.

$$\frac{13}{4}x^{2}-13x+4+9=1$$

Add 4 and 9 to get 13.

$$\frac{13}{4}x^{2}-13x+13=1$$

Subtract 13 from both sides.

$$\frac{13}{4}x^{2}-13x=1-13$$

Subtract 13 from 1 to get -12.

$$\frac{13}{4}x^{2}-13x=-12$$

Divide both sides of the equation by \frac{13}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{13}{4}x^{2}-13x}{\frac{13}{4}}=-\frac{12}{\frac{13}{4}}$$

Dividing by \frac{13}{4} undoes the multiplication by \frac{13}{4}.

$$x^{2}+\left(-\frac{13}{\frac{13}{4}}\right)x=-\frac{12}{\frac{13}{4}}$$

Divide -13 by \frac{13}{4} by multiplying -13 by the reciprocal of \frac{13}{4}.

$$x^{2}-4x=-\frac{12}{\frac{13}{4}}$$

Divide -12 by \frac{13}{4} by multiplying -12 by the reciprocal of \frac{13}{4}.

$$x^{2}-4x=-\frac{48}{13}$$

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

$$x^{2}-4x+\left(-2\right)^{2}=-\frac{48}{13}+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-\frac{48}{13}+4$$

Add -\frac{48}{13} to 4.

$$x^{2}-4x+4=\frac{4}{13}$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=\frac{4}{13}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{4}{13}}$$

Simplify.

$$x-2=\frac{2\sqrt{13}}{13} x-2=-\frac{2\sqrt{13}}{13}$$

Add 2 to both sides of the equation.

$$x=\frac{2\sqrt{13}}{13}+2 x=-\frac{2\sqrt{13}}{13}+2$$