Solve (2sqrt{x-3})^2=(3-x)^2

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

Solve for x

$$x=3$$

$$x=7$$

Steps Using Factoring By Grouping

Expand \left(2\sqrt{x-3}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Calculate \sqrt{x-3} to the power of 2 and get x-3.

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

Use the distributive property to multiply 4 by x-3.

$$4x-12=\left(3-x\right)^{2}$$

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

$$4x-12=9-6x+x^{2}$$

Subtract 9 from both sides.

$$4x-12-9=-6x+x^{2}$$

Subtract 9 from -12 to get -21.

$$4x-21=-6x+x^{2}$$

Add 6x to both sides.

$$4x-21+6x=x^{2}$$

Combine 4x and 6x to get 10x.

$$10x-21=x^{2}$$

Subtract x^{2} from both sides.

$$10x-21-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+10x-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=10 ab=-\left(-21\right)=21$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 21.

$$1,21 3,7$$

Calculate the sum for each pair.

$$1+21=22 3+7=10$$

The solution is the pair that gives sum 10.

$$a=7 b=3$$

Rewrite -x^{2}+10x-21 as \left(-x^{2}+7x\right)+\left(3x-21\right).

$$\left(-x^{2}+7x\right)+\left(3x-21\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-7\right)+3\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(-x+3\right)$$

To find equation solutions, solve x-7=0 and -x+3=0.

$$x=7 x=3$$

Steps Using the Quadratic Formula

Expand \left(2\sqrt{x-3}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Calculate \sqrt{x-3} to the power of 2 and get x-3.

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

Use the distributive property to multiply 4 by x-3.

$$4x-12=\left(3-x\right)^{2}$$

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

$$4x-12=9-6x+x^{2}$$

Subtract 9 from both sides.

$$4x-12-9=-6x+x^{2}$$

Subtract 9 from -12 to get -21.

$$4x-21=-6x+x^{2}$$

Add 6x to both sides.

$$4x-21+6x=x^{2}$$

Combine 4x and 6x to get 10x.

$$10x-21=x^{2}$$

Subtract x^{2} from both sides.

$$10x-21-x^{2}=0$$

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.

$$-x^{2}+10x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-21\right)}}{2\left(-1\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-21\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-10±\sqrt{100+4\left(-21\right)}}{2\left(-1\right)}$$

Multiply 4 times -21.

$$x=\frac{-10±\sqrt{100-84}}{2\left(-1\right)}$$

Add 100 to -84.

$$x=\frac{-10±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$x=\frac{-10±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-10±4}{-2}$$

Now solve the equation x=\frac{-10±4}{-2} when ± is plus. Add -10 to 4.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

Now solve the equation x=\frac{-10±4}{-2} when ± is minus. Subtract 4 from -10.

$$x=-\frac{14}{-2}$$

Divide -14 by -2.

$$x=7$$

The equation is now solved.

$$x=3 x=7$$

Steps for Completing the Square

Expand \left(2\sqrt{x-3}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Calculate \sqrt{x-3} to the power of 2 and get x-3.

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

Use the distributive property to multiply 4 by x-3.

$$4x-12=\left(3-x\right)^{2}$$

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

$$4x-12=9-6x+x^{2}$$

Add 6x to both sides.

$$4x-12+6x=9+x^{2}$$

Combine 4x and 6x to get 10x.

$$10x-12=9+x^{2}$$

Subtract x^{2} from both sides.

$$10x-12-x^{2}=9$$

Add 12 to both sides.

$$10x-x^{2}=9+12$$

Add 9 and 12 to get 21.

$$10x-x^{2}=21$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+10x=21$$

Divide both sides by -1.

$$\frac{-x^{2}+10x}{-1}=\frac{21}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{10}{-1}x=\frac{21}{-1}$$

Divide 10 by -1.

$$x^{2}-10x=\frac{21}{-1}$$

Divide 21 by -1.

$$x^{2}-10x=-21$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-21+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-21+25$$

Add -21 to 25.

$$x^{2}-10x+25=4$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-5=2 x-5=-2$$

Add 5 to both sides of the equation.

$$x=7 x=3$$