Solve (2x+1)^2+(x+1)^2=6x+47

$$( 2 x + 1 ) ^ { 2 } + ( x + 1 ) ^ { 2 } = 6 x + 47$$

Solve for x

$$x=-3$$

$$x=3$$

Steps Using Difference of Squares

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

$$4x^{2}+4x+1+\left(x+1\right)^{2}=6x+47$$

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

$$4x^{2}+4x+1+x^{2}+2x+1=6x+47$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}+4x+1+2x+1=6x+47$$

Combine 4x and 2x to get 6x.

$$5x^{2}+6x+1+1=6x+47$$

Add 1 and 1 to get 2.

$$5x^{2}+6x+2=6x+47$$

Subtract 6x from both sides.

$$5x^{2}+6x+2-6x=47$$

Combine 6x and -6x to get 0.

$$5x^{2}+2=47$$

Subtract 47 from both sides.

$$5x^{2}+2-47=0$$

Subtract 47 from 2 to get -45.

$$5x^{2}-45=0$$

Divide both sides by 5.

$$x^{2}-9=0$$

Consider x^{2}-9. Rewrite x^{2}-9 as x^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-3\right)\left(x+3\right)=0$$

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

$$x=3 x=-3$$

Steps by Finding Square Root

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

$$4x^{2}+4x+1+\left(x+1\right)^{2}=6x+47$$

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

$$4x^{2}+4x+1+x^{2}+2x+1=6x+47$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}+4x+1+2x+1=6x+47$$

Combine 4x and 2x to get 6x.

$$5x^{2}+6x+1+1=6x+47$$

Add 1 and 1 to get 2.

$$5x^{2}+6x+2=6x+47$$

Subtract 6x from both sides.

$$5x^{2}+6x+2-6x=47$$

Combine 6x and -6x to get 0.

$$5x^{2}+2=47$$

Subtract 2 from both sides.

$$5x^{2}=47-2$$

Subtract 2 from 47 to get 45.

$$5x^{2}=45$$

Divide both sides by 5.

$$x^{2}=\frac{45}{5}$$

Divide 45 by 5 to get 9.

$$x^{2}=9$$

Take the square root of both sides of the equation.

$$x=3 x=-3$$

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}.

$$4x^{2}+4x+1+\left(x+1\right)^{2}=6x+47$$

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

$$4x^{2}+4x+1+x^{2}+2x+1=6x+47$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}+4x+1+2x+1=6x+47$$

Combine 4x and 2x to get 6x.

$$5x^{2}+6x+1+1=6x+47$$

Add 1 and 1 to get 2.

$$5x^{2}+6x+2=6x+47$$

Subtract 6x from both sides.

$$5x^{2}+6x+2-6x=47$$

Combine 6x and -6x to get 0.

$$5x^{2}+2=47$$

Subtract 47 from both sides.

$$5x^{2}+2-47=0$$

Subtract 47 from 2 to get -45.

$$5x^{2}-45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 0 for b, and -45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 5\left(-45\right)}}{2\times 5}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 5\left(-45\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{0±\sqrt{-20\left(-45\right)}}{2\times 5}$$

Multiply -20 times -45.

$$x=\frac{0±\sqrt{900}}{2\times 5}$$

Take the square root of 900.

$$x=\frac{0±30}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{0±30}{10}$$

Now solve the equation x=\frac{0±30}{10} when ± is plus. Divide 30 by 10.

$$x=3$$

Now solve the equation x=\frac{0±30}{10} when ± is minus. Divide -30 by 10.

$$x=-3$$

The equation is now solved.

$$x=3 x=-3$$