Solve (x-3)^2=x/4

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

Solve for x

$$x=4$$

$$x = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4.

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

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

$$4\left(x^{2}-6x+9\right)=x$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36=x$$

Subtract x from both sides.

$$4x^{2}-24x+36-x=0$$

Combine -24x and -x to get -25x.

$$4x^{2}-25x+36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+36. To find a and b, set up a system to be solved.

$$a+b=-25 ab=4\times 36=144$$

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 144.

$$-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12$$

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -25.

$$a=-16 b=-9$$

Rewrite 4x^{2}-25x+36 as \left(4x^{2}-16x\right)+\left(-9x+36\right).

$$\left(4x^{2}-16x\right)+\left(-9x+36\right)$$

Factor out 4x in the first and -9 in the second group.

$$4x\left(x-4\right)-9\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(4x-9\right)$$

To find equation solutions, solve x-4=0 and 4x-9=0.

$$x=4 x=\frac{9}{4}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

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

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

$$4\left(x^{2}-6x+9\right)=x$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36=x$$

Subtract x from both sides.

$$4x^{2}-24x+36-x=0$$

Combine -24x and -x to get -25x.

$$4x^{2}-25x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -25 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 4\times 36}}{2\times 4}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 4\times 36}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-25\right)±\sqrt{625-16\times 36}}{2\times 4}$$

Multiply -16 times 36.

$$x=\frac{-\left(-25\right)±\sqrt{625-576}}{2\times 4}$$

Add 625 to -576.

$$x=\frac{-\left(-25\right)±\sqrt{49}}{2\times 4}$$

Take the square root of 49.

$$x=\frac{-\left(-25\right)±7}{2\times 4}$$

The opposite of -25 is 25.

$$x=\frac{25±7}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{25±7}{8}$$

Now solve the equation x=\frac{25±7}{8} when ± is plus. Add 25 to 7.

$$x=\frac{32}{8}$$

Divide 32 by 8.

$$x=4$$

Now solve the equation x=\frac{25±7}{8} when ± is minus. Subtract 7 from 25.

$$x=\frac{18}{8}$$

Reduce the fraction \frac{18}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{4}$$

The equation is now solved.

$$x=4 x=\frac{9}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

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

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

$$4\left(x^{2}-6x+9\right)=x$$

Use the distributive property to multiply 4 by x^{2}-6x+9.

$$4x^{2}-24x+36=x$$

Subtract x from both sides.

$$4x^{2}-24x+36-x=0$$

Combine -24x and -x to get -25x.

$$4x^{2}-25x+36=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-25x=-36$$

Divide both sides by 4.

$$\frac{4x^{2}-25x}{4}=-\frac{36}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{25}{4}x=-\frac{36}{4}$$

Divide -36 by 4.

$$x^{2}-\frac{25}{4}x=-9$$

Divide -\frac{25}{4}, the coefficient of the x term, by 2 to get -\frac{25}{8}. Then add the square of -\frac{25}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{25}{4}x+\left(-\frac{25}{8}\right)^{2}=-9+\left(-\frac{25}{8}\right)^{2}$$

Square -\frac{25}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{25}{4}x+\frac{625}{64}=-9+\frac{625}{64}$$

Add -9 to \frac{625}{64}.

$$x^{2}-\frac{25}{4}x+\frac{625}{64}=\frac{49}{64}$$

Factor x^{2}-\frac{25}{4}x+\frac{625}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{25}{8}\right)^{2}=\frac{49}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Simplify.

$$x-\frac{25}{8}=\frac{7}{8} x-\frac{25}{8}=-\frac{7}{8}$$

Add \frac{25}{8} to both sides of the equation.

$$x=4 x=\frac{9}{4}$$