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

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

Solve for x

$$x=2$$

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

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

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

$$2\left(x^{2}-6x+9\right)=x$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$2x^{2}-12x+18=x$$

Subtract x from both sides.

$$2x^{2}-12x+18-x=0$$

Combine -12x and -x to get -13x.

$$2x^{2}-13x+18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+18. To find a and b, set up a system to be solved.

$$a+b=-13 ab=2\times 18=36$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -13.

$$a=-9 b=-4$$

Rewrite 2x^{2}-13x+18 as \left(2x^{2}-9x\right)+\left(-4x+18\right).

$$\left(2x^{2}-9x\right)+\left(-4x+18\right)$$

Factor out x in the first and -2 in the second group.

$$x\left(2x-9\right)-2\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(x-2\right)$$

To find equation solutions, solve 2x-9=0 and x-2=0.

$$x=\frac{9}{2} x=2$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

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

$$2\left(x^{2}-6x+9\right)=x$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$2x^{2}-12x+18=x$$

Subtract x from both sides.

$$2x^{2}-12x+18-x=0$$

Combine -12x and -x to get -13x.

$$2x^{2}-13x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -13 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2\times 18}}{2\times 2}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 2\times 18}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-13\right)±\sqrt{169-8\times 18}}{2\times 2}$$

Multiply -8 times 18.

$$x=\frac{-\left(-13\right)±\sqrt{169-144}}{2\times 2}$$

Add 169 to -144.

$$x=\frac{-\left(-13\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-\left(-13\right)±5}{2\times 2}$$

The opposite of -13 is 13.

$$x=\frac{13±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{13±5}{4}$$

Now solve the equation x=\frac{13±5}{4} when ± is plus. Add 13 to 5.

$$x=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{13±5}{4} when ± is minus. Subtract 5 from 13.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

The equation is now solved.

$$x=\frac{9}{2} x=2$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

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

$$2\left(x^{2}-6x+9\right)=x$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$2x^{2}-12x+18=x$$

Subtract x from both sides.

$$2x^{2}-12x+18-x=0$$

Combine -12x and -x to get -13x.

$$2x^{2}-13x+18=0$$

Subtract 18 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-13x=-18$$

Divide both sides by 2.

$$\frac{2x^{2}-13x}{2}=-\frac{18}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{13}{2}x=-\frac{18}{2}$$

Divide -18 by 2.

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

Divide -\frac{13}{2}, the coefficient of the x term, by 2 to get -\frac{13}{4}. Then add the square of -\frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=-9+\left(-\frac{13}{4}\right)^{2}$$

Square -\frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=-9+\frac{169}{16}$$

Add -9 to \frac{169}{16}.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{13}{2}x+\frac{169}{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}.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{5}{4} x-\frac{13}{4}=-\frac{5}{4}$$

Add \frac{13}{4} to both sides of the equation.

$$x=\frac{9}{2} x=2$$