Solve (2x+1)^2+(2x+3)^2=34

$$( 2 x + 1 ) ^ { 2 } + ( 2 x + 3 ) ^ { 2 } = 34$$

Solve for x

$$x=-3$$

$$x=1$$

Steps Using Factoring By Grouping

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

$$4x^{2}+4x+1+\left(2x+3\right)^{2}=34$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}+4x+1+4x^{2}+12x+9=34$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}+4x+1+12x+9=34$$

Combine 4x and 12x to get 16x.

$$8x^{2}+16x+1+9=34$$

Add 1 and 9 to get 10.

$$8x^{2}+16x+10=34$$

Subtract 34 from both sides.

$$8x^{2}+16x+10-34=0$$

Subtract 34 from 10 to get -24.

$$8x^{2}+16x-24=0$$

Divide both sides by 8.

$$x^{2}+2x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-1 b=3$$

Rewrite x^{2}+2x-3 as \left(x^{2}-x\right)+\left(3x-3\right).

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

Factor out x in the first and 3 in the second group.

$$x\left(x-1\right)+3\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+3\right)$$

To find equation solutions, solve x-1=0 and x+3=0.

$$x=1 x=-3$$

Steps Using the Quadratic Formula

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

$$4x^{2}+4x+1+\left(2x+3\right)^{2}=34$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}+4x+1+4x^{2}+12x+9=34$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}+4x+1+12x+9=34$$

Combine 4x and 12x to get 16x.

$$8x^{2}+16x+1+9=34$$

Add 1 and 9 to get 10.

$$8x^{2}+16x+10=34$$

Subtract 34 from both sides.

$$8x^{2}+16x+10-34=0$$

Subtract 34 from 10 to get -24.

$$8x^{2}+16x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 16 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 8\left(-24\right)}}{2\times 8}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 8\left(-24\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-16±\sqrt{256-32\left(-24\right)}}{2\times 8}$$

Multiply -32 times -24.

$$x=\frac{-16±\sqrt{256+768}}{2\times 8}$$

Add 256 to 768.

$$x=\frac{-16±\sqrt{1024}}{2\times 8}$$

Take the square root of 1024.

$$x=\frac{-16±32}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-16±32}{16}$$

Now solve the equation x=\frac{-16±32}{16} when ± is plus. Add -16 to 32.

$$x=\frac{16}{16}$$

Divide 16 by 16.

$$x=1$$

Now solve the equation x=\frac{-16±32}{16} when ± is minus. Subtract 32 from -16.

$$x=-\frac{48}{16}$$

Divide -48 by 16.

$$x=-3$$

The equation is now solved.

$$x=1 x=-3$$

Steps for Completing the Square

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

$$4x^{2}+4x+1+\left(2x+3\right)^{2}=34$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}+4x+1+4x^{2}+12x+9=34$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}+4x+1+12x+9=34$$

Combine 4x and 12x to get 16x.

$$8x^{2}+16x+1+9=34$$

Add 1 and 9 to get 10.

$$8x^{2}+16x+10=34$$

Subtract 10 from both sides.

$$8x^{2}+16x=34-10$$

Subtract 10 from 34 to get 24.

$$8x^{2}+16x=24$$

Divide both sides by 8.

$$\frac{8x^{2}+16x}{8}=\frac{24}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{16}{8}x=\frac{24}{8}$$

Divide 16 by 8.

$$x^{2}+2x=\frac{24}{8}$$

Divide 24 by 8.

$$x^{2}+2x=3$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square 1.

$$x^{2}+2x+1=3+1$$

Add 3 to 1.

$$x^{2}+2x+1=4$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+1=2 x+1=-2$$

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$