Solve left(x+8right)^2=left(x+4right)^2+x^2

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

Solve for x

$$x=-4$$

$$x=12$$

Steps Using Factoring By Grouping

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

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

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

$$x^{2}+16x+64=x^{2}+8x+16+x^{2}$$

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

$$x^{2}+16x+64=2x^{2}+8x+16$$

Subtract 2x^{2} from both sides.

$$x^{2}+16x+64-2x^{2}=8x+16$$

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

$$-x^{2}+16x+64=8x+16$$

Subtract 8x from both sides.

$$-x^{2}+16x+64-8x=16$$

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

$$-x^{2}+8x+64=16$$

Subtract 16 from both sides.

$$-x^{2}+8x+64-16=0$$

Subtract 16 from 64 to get 48.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+48. To find a and b, set up a system to be solved.

$$a+b=8 ab=-48=-48$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=12 b=-4$$

Rewrite -x^{2}+8x+48 as \left(-x^{2}+12x\right)+\left(-4x+48\right).

$$\left(-x^{2}+12x\right)+\left(-4x+48\right)$$

Factor out -x in the first and -4 in the second group.

$$-x\left(x-12\right)-4\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(-x-4\right)$$

To find equation solutions, solve x-12=0 and -x-4=0.

$$x=12 x=-4$$

Steps Using the Quadratic Formula

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

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

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

$$x^{2}+16x+64=x^{2}+8x+16+x^{2}$$

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

$$x^{2}+16x+64=2x^{2}+8x+16$$

Subtract 2x^{2} from both sides.

$$x^{2}+16x+64-2x^{2}=8x+16$$

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

$$-x^{2}+16x+64=8x+16$$

Subtract 8x from both sides.

$$-x^{2}+16x+64-8x=16$$

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

$$-x^{2}+8x+64=16$$

Subtract 16 from both sides.

$$-x^{2}+8x+64-16=0$$

Subtract 16 from 64 to get 48.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\times 48}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\times 48}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\times 48}}{2\left(-1\right)}$$

Multiply 4 times 48.

$$x=\frac{-8±\sqrt{64+192}}{2\left(-1\right)}$$

Add 64 to 192.

$$x=\frac{-8±\sqrt{256}}{2\left(-1\right)}$$

Take the square root of 256.

$$x=\frac{-8±16}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±16}{-2}$$

Now solve the equation x=\frac{-8±16}{-2} when ± is plus. Add -8 to 16.

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

Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{-8±16}{-2} when ± is minus. Subtract 16 from -8.

$$x=-\frac{24}{-2}$$

Divide -24 by -2.

$$x=12$$

The equation is now solved.

$$x=-4 x=12$$

Steps for Completing the Square

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

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

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

$$x^{2}+16x+64=x^{2}+8x+16+x^{2}$$

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

$$x^{2}+16x+64=2x^{2}+8x+16$$

Subtract 2x^{2} from both sides.

$$x^{2}+16x+64-2x^{2}=8x+16$$

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

$$-x^{2}+16x+64=8x+16$$

Subtract 8x from both sides.

$$-x^{2}+16x+64-8x=16$$

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

$$-x^{2}+8x+64=16$$

Subtract 64 from both sides.

$$-x^{2}+8x=16-64$$

Subtract 64 from 16 to get -48.

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

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=-\frac{48}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=-\frac{48}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=-\frac{48}{-1}$$

Divide -48 by -1.

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

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -4.

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

Add 48 to 16.

$$x^{2}-8x+16=64$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{64}$$

Simplify.

$$x-4=8 x-4=-8$$

Add 4 to both sides of the equation.

$$x=12 x=-4$$