Solve 1/4left(4^2+left(8-xright)^2+left(4+xright)^2right)=88

$$\frac{ 1 }{ 4 } \left( { 4 }^{ 2 } + { \left(8-x \right) }^{ 2 } + { \left(4+x \right) }^{ 2 } \right) = 88$$

Solve for x

$$x=2\sqrt{33}+2\approx 13.489125293$$

$$x=2-2\sqrt{33}\approx -9.489125293$$

Steps Using the Quadratic Formula

Multiply both sides by 4, the reciprocal of \frac{1}{4}.

$$4^{2}+\left(8-x\right)^{2}+\left(4+x\right)^{2}=88\times 4$$

Multiply 88 and 4 to get 352.

$$4^{2}+\left(8-x\right)^{2}+\left(4+x\right)^{2}=352$$

Calculate 4 to the power of 2 and get 16.

$$16+\left(8-x\right)^{2}+\left(4+x\right)^{2}=352$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-x\right)^{2}.

$$16+64-16x+x^{2}+\left(4+x\right)^{2}=352$$

Add 16 and 64 to get 80.

$$80-16x+x^{2}+\left(4+x\right)^{2}=352$$

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

$$80-16x+x^{2}+16+8x+x^{2}=352$$

Add 80 and 16 to get 96.

$$96-16x+x^{2}+8x+x^{2}=352$$

Combine -16x and 8x to get -8x.

$$96-8x+x^{2}+x^{2}=352$$

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

$$96-8x+2x^{2}=352$$

Subtract 352 from both sides.

$$96-8x+2x^{2}-352=0$$

Subtract 352 from 96 to get -256.

$$-256-8x+2x^{2}=0$$

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.

$$2x^{2}-8x-256=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and -256 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-256\right)}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-256\right)}}{2\times 2}$$

Multiply -4 times 2.

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

Multiply -8 times -256.

$$x=\frac{-\left(-8\right)±\sqrt{64+2048}}{2\times 2}$$

Add 64 to 2048.

$$x=\frac{-\left(-8\right)±\sqrt{2112}}{2\times 2}$$

Take the square root of 2112.

$$x=\frac{-\left(-8\right)±8\sqrt{33}}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±8\sqrt{33}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±8\sqrt{33}}{4}$$

Now solve the equation x=\frac{8±8\sqrt{33}}{4} when ± is plus. Add 8 to 8\sqrt{33}.

$$x=\frac{8\sqrt{33}+8}{4}$$

Divide 8+8\sqrt{33} by 4.

$$x=2\sqrt{33}+2$$

Now solve the equation x=\frac{8±8\sqrt{33}}{4} when ± is minus. Subtract 8\sqrt{33} from 8.

$$x=\frac{8-8\sqrt{33}}{4}$$

Divide 8-8\sqrt{33} by 4.

$$x=2-2\sqrt{33}$$

The equation is now solved.

$$x=2\sqrt{33}+2 x=2-2\sqrt{33}$$

Steps for Completing the Square

Multiply both sides by 4, the reciprocal of \frac{1}{4}.

$$4^{2}+\left(8-x\right)^{2}+\left(4+x\right)^{2}=88\times 4$$

Multiply 88 and 4 to get 352.

$$4^{2}+\left(8-x\right)^{2}+\left(4+x\right)^{2}=352$$

Calculate 4 to the power of 2 and get 16.

$$16+\left(8-x\right)^{2}+\left(4+x\right)^{2}=352$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-x\right)^{2}.

$$16+64-16x+x^{2}+\left(4+x\right)^{2}=352$$

Add 16 and 64 to get 80.

$$80-16x+x^{2}+\left(4+x\right)^{2}=352$$

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

$$80-16x+x^{2}+16+8x+x^{2}=352$$

Add 80 and 16 to get 96.

$$96-16x+x^{2}+8x+x^{2}=352$$

Combine -16x and 8x to get -8x.

$$96-8x+x^{2}+x^{2}=352$$

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

$$96-8x+2x^{2}=352$$

Subtract 96 from both sides.

$$-8x+2x^{2}=352-96$$

Subtract 96 from 352 to get 256.

$$-8x+2x^{2}=256$$

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.

$$2x^{2}-8x=256$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=\frac{256}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=\frac{256}{2}$$

Divide -8 by 2.

$$x^{2}-4x=\frac{256}{2}$$

Divide 256 by 2.

$$x^{2}-4x=128$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=128+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=128+4$$

Add 128 to 4.

$$x^{2}-4x+4=132$$

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}.

$$\left(x-2\right)^{2}=132$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{132}$$

Simplify.

$$x-2=2\sqrt{33} x-2=-2\sqrt{33}$$

Add 2 to both sides of the equation.

$$x=2\sqrt{33}+2 x=2-2\sqrt{33}$$