Solve 16+x^2+left(4-xright)^2+16={left(4sqrt{5}right)}^2

$$16+ { x }^{ 2 } + { \left(4-x \right) }^{ 2 } +16= { \left(4 \sqrt{ 5 } \right) }^{ 2 }$$

Solve for x

$$x=2\sqrt{5}+2\approx 6.472135955$$

$$x=2-2\sqrt{5}\approx -2.472135955$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$16+x^{2}+16-8x+x^{2}+16=\left(4\sqrt{5}\right)^{2}$$

Add 16 and 16 to get 32.

$$32+x^{2}-8x+x^{2}+16=\left(4\sqrt{5}\right)^{2}$$

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

$$32+2x^{2}-8x+16=\left(4\sqrt{5}\right)^{2}$$

Add 32 and 16 to get 48.

$$48+2x^{2}-8x=\left(4\sqrt{5}\right)^{2}$$

Expand \left(4\sqrt{5}\right)^{2}.

$$48+2x^{2}-8x=4^{2}\left(\sqrt{5}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$48+2x^{2}-8x=16\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$48+2x^{2}-8x=16\times 5$$

Multiply 16 and 5 to get 80.

$$48+2x^{2}-8x=80$$

Subtract 80 from both sides.

$$48+2x^{2}-8x-80=0$$

Subtract 80 from 48 to get -32.

$$-32+2x^{2}-8x=0$$

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.

$$2x^{2}-8x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-32\right)}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-32\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\left(-32\right)}}{2\times 2}$$

Multiply -8 times -32.

$$x=\frac{-\left(-8\right)±\sqrt{64+256}}{2\times 2}$$

Add 64 to 256.

$$x=\frac{-\left(-8\right)±\sqrt{320}}{2\times 2}$$

Take the square root of 320.

$$x=\frac{-\left(-8\right)±8\sqrt{5}}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±8\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±8\sqrt{5}}{4}$$

Now solve the equation x=\frac{8±8\sqrt{5}}{4} when ± is plus. Add 8 to 8\sqrt{5}.

$$x=\frac{8\sqrt{5}+8}{4}$$

Divide 8+8\sqrt{5} by 4.

$$x=2\sqrt{5}+2$$

Now solve the equation x=\frac{8±8\sqrt{5}}{4} when ± is minus. Subtract 8\sqrt{5} from 8.

$$x=\frac{8-8\sqrt{5}}{4}$$

Divide 8-8\sqrt{5} by 4.

$$x=2-2\sqrt{5}$$

The equation is now solved.

$$x=2\sqrt{5}+2 x=2-2\sqrt{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$16+x^{2}+16-8x+x^{2}+16=\left(4\sqrt{5}\right)^{2}$$

Add 16 and 16 to get 32.

$$32+x^{2}-8x+x^{2}+16=\left(4\sqrt{5}\right)^{2}$$

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

$$32+2x^{2}-8x+16=\left(4\sqrt{5}\right)^{2}$$

Add 32 and 16 to get 48.

$$48+2x^{2}-8x=\left(4\sqrt{5}\right)^{2}$$

Expand \left(4\sqrt{5}\right)^{2}.

$$48+2x^{2}-8x=4^{2}\left(\sqrt{5}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$48+2x^{2}-8x=16\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$48+2x^{2}-8x=16\times 5$$

Multiply 16 and 5 to get 80.

$$48+2x^{2}-8x=80$$

Subtract 48 from both sides.

$$2x^{2}-8x=80-48$$

Subtract 48 from 80 to get 32.

$$2x^{2}-8x=32$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=\frac{32}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=\frac{32}{2}$$

Divide -8 by 2.

$$x^{2}-4x=\frac{32}{2}$$

Divide 32 by 2.

$$x^{2}-4x=16$$

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

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

Square -2.

$$x^{2}-4x+4=16+4$$

Add 16 to 4.

$$x^{2}-4x+4=20$$

Factor x^{2}-4x+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}.

$$\left(x-2\right)^{2}=20$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{20}$$

Simplify.

$$x-2=2\sqrt{5} x-2=-2\sqrt{5}$$

Add 2 to both sides of the equation.

$$x=2\sqrt{5}+2 x=2-2\sqrt{5}$$