Solve x^2+left(9-xright)^2=6^2

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

Solve for x (complex solution)

$$x=\frac{9}{2}+\frac{3}{2}i=4.5+1.5i$$

$$x=\frac{9}{2}-\frac{3}{2}i=4.5-1.5i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$x^{2}+81-18x+x^{2}=6^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+81-18x=6^{2}$$

Calculate 6 to the power of 2 and get 36.

$$2x^{2}+81-18x=36$$

Subtract 36 from both sides.

$$2x^{2}+81-18x-36=0$$

Subtract 36 from 81 to get 45.

$$2x^{2}+45-18x=0$$

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.

$$2x^{2}-18x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -18 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 2\times 45}}{2\times 2}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 2\times 45}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-18\right)±\sqrt{324-8\times 45}}{2\times 2}$$

Multiply -8 times 45.

$$x=\frac{-\left(-18\right)±\sqrt{324-360}}{2\times 2}$$

Add 324 to -360.

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

Take the square root of -36.

$$x=\frac{-\left(-18\right)±6i}{2\times 2}$$

The opposite of -18 is 18.

$$x=\frac{18±6i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{18±6i}{4}$$

Now solve the equation x=\frac{18±6i}{4} when ± is plus. Add 18 to 6i.

$$x=\frac{18+6i}{4}$$

Divide 18+6i by 4.

$$x=\frac{9}{2}+\frac{3}{2}i$$

Now solve the equation x=\frac{18±6i}{4} when ± is minus. Subtract 6i from 18.

$$x=\frac{18-6i}{4}$$

Divide 18-6i by 4.

$$x=\frac{9}{2}-\frac{3}{2}i$$

The equation is now solved.

$$x=\frac{9}{2}+\frac{3}{2}i x=\frac{9}{2}-\frac{3}{2}i$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-x\right)^{2}.

$$x^{2}+81-18x+x^{2}=6^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+81-18x=6^{2}$$

Calculate 6 to the power of 2 and get 36.

$$2x^{2}+81-18x=36$$

Subtract 81 from both sides.

$$2x^{2}-18x=36-81$$

Subtract 81 from 36 to get -45.

$$2x^{2}-18x=-45$$

Divide both sides by 2.

$$\frac{2x^{2}-18x}{2}=-\frac{45}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{18}{2}\right)x=-\frac{45}{2}$$

Divide -18 by 2.

$$x^{2}-9x=-\frac{45}{2}$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-\frac{45}{2}+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-\frac{45}{2}+\frac{81}{4}$$

Add -\frac{45}{2} to \frac{81}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-9x+\frac{81}{4}=-\frac{9}{4}$$

Factor x^{2}-9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{9}{2}=\frac{3}{2}i x-\frac{9}{2}=-\frac{3}{2}i$$

Add \frac{9}{2} to both sides of the equation.

$$x=\frac{9}{2}+\frac{3}{2}i x=\frac{9}{2}-\frac{3}{2}i$$