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

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

Solve for x

$$x=3-\sqrt{6}\approx 0.550510257$$

$$x=\sqrt{6}+3\approx 5.449489743$$

Steps Using the Quadratic Formula

Calculate 3 to the power of 2 and get 9.

$$9=\left(\sqrt{3}\right)^{2}+\left(3-x\right)^{2}$$

The square of \sqrt{3} is 3.

$$9=3+\left(3-x\right)^{2}$$

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

$$9=3+9-6x+x^{2}$$

Add 3 and 9 to get 12.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 9 from both sides.

$$12-6x+x^{2}-9=0$$

Subtract 9 from 12 to get 3.

$$3-6x+x^{2}=0$$

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.

$$x^{2}-6x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 3}}{2}$$

Multiply -4 times 3.

$$x=\frac{-\left(-6\right)±\sqrt{36-12}}{2}$$

Add 36 to -12.

$$x=\frac{-\left(-6\right)±\sqrt{24}}{2}$$

Take the square root of 24.

$$x=\frac{-\left(-6\right)±2\sqrt{6}}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{6}}{2}$$

Now solve the equation x=\frac{6±2\sqrt{6}}{2} when ± is plus. Add 6 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}+6}{2}$$

Divide 6+2\sqrt{6} by 2.

$$x=\sqrt{6}+3$$

Now solve the equation x=\frac{6±2\sqrt{6}}{2} when ± is minus. Subtract 2\sqrt{6} from 6.

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

Divide 6-2\sqrt{6} by 2.

$$x=3-\sqrt{6}$$

The equation is now solved.

$$x=\sqrt{6}+3 x=3-\sqrt{6}$$

Steps for Completing the Square

Calculate 3 to the power of 2 and get 9.

$$9=\left(\sqrt{3}\right)^{2}+\left(3-x\right)^{2}$$

The square of \sqrt{3} is 3.

$$9=3+\left(3-x\right)^{2}$$

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

$$9=3+9-6x+x^{2}$$

Add 3 and 9 to get 12.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 12 from both sides.

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

Subtract 12 from 9 to get -3.

$$-6x+x^{2}=-3$$

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.

$$x^{2}-6x=-3$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-3+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-3+9$$

Add -3 to 9.

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

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=6$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{6}$$

Simplify.

$$x-3=\sqrt{6} x-3=-\sqrt{6}$$

Add 3 to both sides of the equation.

$$x=\sqrt{6}+3 x=3-\sqrt{6}$$