Solve (x-7)^2+x^2=(x+1)^2

$$( x - 7 ) ^ { 2 } + x ^ { 2 } = ( x + 1 ) ^ { 2 }$$

Solve for x

$$x=4$$

$$x=12$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x^{2}-14x+49+x^{2}=\left(x+1\right)^{2}$$

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

$$2x^{2}-14x+49=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$2x^{2}-14x+49=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$2x^{2}-14x+49-x^{2}=2x+1$$

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

$$x^{2}-14x+49=2x+1$$

Subtract 2x from both sides.

$$x^{2}-14x+49-2x=1$$

Combine -14x and -2x to get -16x.

$$x^{2}-16x+49=1$$

Subtract 1 from both sides.

$$x^{2}-16x+49-1=0$$

Subtract 1 from 49 to get 48.

$$x^{2}-16x+48=0$$

To solve the equation, factor x^{2}-16x+48 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-16 ab=48$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both 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.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -16.

$$a=-12 b=-4$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

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

$$x=12 x=4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x^{2}-14x+49+x^{2}=\left(x+1\right)^{2}$$

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

$$2x^{2}-14x+49=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$2x^{2}-14x+49=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$2x^{2}-14x+49-x^{2}=2x+1$$

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

$$x^{2}-14x+49=2x+1$$

Subtract 2x from both sides.

$$x^{2}-14x+49-2x=1$$

Combine -14x and -2x to get -16x.

$$x^{2}-16x+49=1$$

Subtract 1 from both sides.

$$x^{2}-16x+49-1=0$$

Subtract 1 from 49 to get 48.

$$x^{2}-16x+48=0$$

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.

$$a+b=-16 ab=1\times 48=48$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both 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.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -16.

$$a=-12 b=-4$$

Rewrite x^{2}-16x+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-7\right)^{2}.

$$x^{2}-14x+49+x^{2}=\left(x+1\right)^{2}$$

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

$$2x^{2}-14x+49=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$2x^{2}-14x+49=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$2x^{2}-14x+49-x^{2}=2x+1$$

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

$$x^{2}-14x+49=2x+1$$

Subtract 2x from both sides.

$$x^{2}-14x+49-2x=1$$

Combine -14x and -2x to get -16x.

$$x^{2}-16x+49=1$$

Subtract 1 from both sides.

$$x^{2}-16x+49-1=0$$

Subtract 1 from 49 to get 48.

$$x^{2}-16x+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -16 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 48}}{2}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 48}}{2}$$

Multiply -4 times 48.

$$x=\frac{-\left(-16\right)±\sqrt{256-192}}{2}$$

Add 256 to -192.

$$x=\frac{-\left(-16\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-\left(-16\right)±8}{2}$$

The opposite of -16 is 16.

$$x=\frac{16±8}{2}$$

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

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

Divide 24 by 2.

$$x=12$$

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

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

Divide 8 by 2.

$$x=4$$

The equation is now solved.

$$x=12 x=4$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x^{2}-14x+49+x^{2}=\left(x+1\right)^{2}$$

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

$$2x^{2}-14x+49=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$2x^{2}-14x+49=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$2x^{2}-14x+49-x^{2}=2x+1$$

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

$$x^{2}-14x+49=2x+1$$

Subtract 2x from both sides.

$$x^{2}-14x+49-2x=1$$

Combine -14x and -2x to get -16x.

$$x^{2}-16x+49=1$$

Subtract 49 from both sides.

$$x^{2}-16x=1-49$$

Subtract 49 from 1 to get -48.

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

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -8.

$$x^{2}-16x+64=-48+64$$

Add -48 to 64.

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

Factor x^{2}-16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 8 to both sides of the equation.

$$x=12 x=4$$