Solve left(x+2right)^2+left(x+4right)^2=10

$${ \left(x+2 \right) }^{ 2 } + { \left(x+4 \right) }^{ 2 } = 10$$

Solve for x

$$x=-5$$

$$x=-1$$

Steps Using Factoring By Grouping

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

$$x^{2}+4x+4+\left(x+4\right)^{2}=10$$

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

$$x^{2}+4x+4+x^{2}+8x+16=10$$

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

$$2x^{2}+4x+4+8x+16=10$$

Combine 4x and 8x to get 12x.

$$2x^{2}+12x+4+16=10$$

Add 4 and 16 to get 20.

$$2x^{2}+12x+20=10$$

Subtract 10 from both sides.

$$2x^{2}+12x+20-10=0$$

Subtract 10 from 20 to get 10.

$$2x^{2}+12x+10=0$$

Divide both sides by 2.

$$x^{2}+6x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\times 5=5$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=5$$

Rewrite x^{2}+6x+5 as \left(x^{2}+x\right)+\left(5x+5\right).

$$\left(x^{2}+x\right)+\left(5x+5\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x+1\right)+5\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+5\right)$$

To find equation solutions, solve x+1=0 and x+5=0.

$$x=-1 x=-5$$

Steps Using the Quadratic Formula

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

$$x^{2}+4x+4+\left(x+4\right)^{2}=10$$

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

$$x^{2}+4x+4+x^{2}+8x+16=10$$

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

$$2x^{2}+4x+4+8x+16=10$$

Combine 4x and 8x to get 12x.

$$2x^{2}+12x+4+16=10$$

Add 4 and 16 to get 20.

$$2x^{2}+12x+20=10$$

Subtract 10 from both sides.

$$2x^{2}+12x+20-10=0$$

Subtract 10 from 20 to get 10.

$$2x^{2}+12x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 12 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 2\times 10}}{2\times 2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 2\times 10}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-12±\sqrt{144-8\times 10}}{2\times 2}$$

Multiply -8 times 10.

$$x=\frac{-12±\sqrt{144-80}}{2\times 2}$$

Add 144 to -80.

$$x=\frac{-12±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$x=\frac{-12±8}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-12±8}{4}$$

Now solve the equation x=\frac{-12±8}{4} when ± is plus. Add -12 to 8.

$$x=-\frac{4}{4}$$

Divide -4 by 4.

$$x=-1$$

Now solve the equation x=\frac{-12±8}{4} when ± is minus. Subtract 8 from -12.

$$x=-\frac{20}{4}$$

Divide -20 by 4.

$$x=-5$$

The equation is now solved.

$$x=-1 x=-5$$

Steps for Completing the Square

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

$$x^{2}+4x+4+\left(x+4\right)^{2}=10$$

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

$$x^{2}+4x+4+x^{2}+8x+16=10$$

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

$$2x^{2}+4x+4+8x+16=10$$

Combine 4x and 8x to get 12x.

$$2x^{2}+12x+4+16=10$$

Add 4 and 16 to get 20.

$$2x^{2}+12x+20=10$$

Subtract 20 from both sides.

$$2x^{2}+12x=10-20$$

Subtract 20 from 10 to get -10.

$$2x^{2}+12x=-10$$

Divide both sides by 2.

$$\frac{2x^{2}+12x}{2}=-\frac{10}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{12}{2}x=-\frac{10}{2}$$

Divide 12 by 2.

$$x^{2}+6x=-\frac{10}{2}$$

Divide -10 by 2.

$$x^{2}+6x=-5$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-5+3^{2}$$

Square 3.

$$x^{2}+6x+9=-5+9$$

Add -5 to 9.

$$x^{2}+6x+9=4$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+3=2 x+3=-2$$

Subtract 3 from both sides of the equation.

$$x=-1 x=-5$$