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

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

Solve for x

$$x=\sqrt{3}+1\approx 2.732050808$$

$$x=1-\sqrt{3}\approx -0.732050808$$

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}.

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

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

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

Use the distributive property to multiply 3 by x^{2}+2x+1.

$$4x^{2}+4x+1=3x^{2}+6x+3$$

Subtract 3x^{2} from both sides.

$$4x^{2}+4x+1-3x^{2}=6x+3$$

Combine 4x^{2} and -3x^{2} to get x^{2}.

$$x^{2}+4x+1=6x+3$$

Subtract 6x from both sides.

$$x^{2}+4x+1-6x=3$$

Combine 4x and -6x to get -2x.

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

Subtract 3 from both sides.

$$x^{2}-2x+1-3=0$$

Subtract 3 from 1 to get -2.

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

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

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

Square -2.

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

Multiply -4 times -2.

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

Add 4 to 8.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{-\left(-2\right)±2\sqrt{3}}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{3}}{2}$$

Now solve the equation x=\frac{2±2\sqrt{3}}{2} when ± is plus. Add 2 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+2}{2}$$

Divide 2+2\sqrt{3} by 2.

$$x=\sqrt{3}+1$$

Now solve the equation x=\frac{2±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 2.

$$x=\frac{2-2\sqrt{3}}{2}$$

Divide 2-2\sqrt{3} by 2.

$$x=1-\sqrt{3}$$

The equation is now solved.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$

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}.

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

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

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

Use the distributive property to multiply 3 by x^{2}+2x+1.

$$4x^{2}+4x+1=3x^{2}+6x+3$$

Subtract 3x^{2} from both sides.

$$4x^{2}+4x+1-3x^{2}=6x+3$$

Combine 4x^{2} and -3x^{2} to get x^{2}.

$$x^{2}+4x+1=6x+3$$

Subtract 6x from both sides.

$$x^{2}+4x+1-6x=3$$

Combine 4x and -6x to get -2x.

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

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}.

$$\left(x-1\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{3}$$

Simplify.

$$x-1=\sqrt{3} x-1=-\sqrt{3}$$

Add 1 to both sides of the equation.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$