Solve {left(frac{sqrt{36}}{2}-xright)}^2+{left(frac{sqrt{36}}{2}right)}^2=4x^2

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

Solve for x

$$x=\sqrt{7}-1\approx 1.645751311$$

$$x=-\left(\sqrt{7}+1\right)\approx -3.645751311$$

Steps Using the Quadratic Formula

Calculate the square root of 36 and get 6.

$$\left(\frac{6}{2}-x\right)^{2}+\left(\frac{\sqrt{36}}{2}\right)^{2}=4x^{2}$$

Divide 6 by 2 to get 3.

$$\left(3-x\right)^{2}+\left(\frac{\sqrt{36}}{2}\right)^{2}=4x^{2}$$

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

$$9-6x+x^{2}+\left(\frac{\sqrt{36}}{2}\right)^{2}=4x^{2}$$

Calculate the square root of 36 and get 6.

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

Divide 6 by 2 to get 3.

$$9-6x+x^{2}+3^{2}=4x^{2}$$

Calculate 3 to the power of 2 and get 9.

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

Add 9 and 9 to get 18.

$$18-6x+x^{2}=4x^{2}$$

Subtract 4x^{2} from both sides.

$$18-6x+x^{2}-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$18-6x-3x^{2}=0$$

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.

$$-3x^{2}-6x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -6 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-3\right)\times 18}}{2\left(-3\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-3\right)\times 18}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-6\right)±\sqrt{36+12\times 18}}{2\left(-3\right)}$$

Multiply 12 times 18.

$$x=\frac{-\left(-6\right)±\sqrt{36+216}}{2\left(-3\right)}$$

Add 36 to 216.

$$x=\frac{-\left(-6\right)±\sqrt{252}}{2\left(-3\right)}$$

Take the square root of 252.

$$x=\frac{-\left(-6\right)±6\sqrt{7}}{2\left(-3\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±6\sqrt{7}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{6±6\sqrt{7}}{-6}$$

Now solve the equation x=\frac{6±6\sqrt{7}}{-6} when ± is plus. Add 6 to 6\sqrt{7}.

$$x=\frac{6\sqrt{7}+6}{-6}$$

Divide 6+6\sqrt{7} by -6.

$$x=-\left(\sqrt{7}+1\right)$$

Now solve the equation x=\frac{6±6\sqrt{7}}{-6} when ± is minus. Subtract 6\sqrt{7} from 6.

$$x=\frac{6-6\sqrt{7}}{-6}$$

Divide 6-6\sqrt{7} by -6.

$$x=\sqrt{7}-1$$

The equation is now solved.

$$x=-\left(\sqrt{7}+1\right) x=\sqrt{7}-1$$

Steps for Completing the Square

Calculate the square root of 36 and get 6.

$$\left(\frac{6}{2}-x\right)^{2}+\left(\frac{\sqrt{36}}{2}\right)^{2}=4x^{2}$$

Divide 6 by 2 to get 3.

$$\left(3-x\right)^{2}+\left(\frac{\sqrt{36}}{2}\right)^{2}=4x^{2}$$

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

$$9-6x+x^{2}+\left(\frac{\sqrt{36}}{2}\right)^{2}=4x^{2}$$

Calculate the square root of 36 and get 6.

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

Divide 6 by 2 to get 3.

$$9-6x+x^{2}+3^{2}=4x^{2}$$

Calculate 3 to the power of 2 and get 9.

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

Add 9 and 9 to get 18.

$$18-6x+x^{2}=4x^{2}$$

Subtract 4x^{2} from both sides.

$$18-6x+x^{2}-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$18-6x-3x^{2}=0$$

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

$$-6x-3x^{2}=-18$$

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.

$$-3x^{2}-6x=-18$$

Divide both sides by -3.

$$\frac{-3x^{2}-6x}{-3}=-\frac{18}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{6}{-3}\right)x=-\frac{18}{-3}$$

Divide -6 by -3.

$$x^{2}+2x=-\frac{18}{-3}$$

Divide -18 by -3.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=6+1^{2}$$

Square 1.

$$x^{2}+2x+1=6+1$$

Add 6 to 1.

$$x^{2}+2x+1=7$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{7}$$

Simplify.

$$x+1=\sqrt{7} x+1=-\sqrt{7}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{7}-1 x=-\sqrt{7}-1$$