Solve (2x+1)^2-x^2=0

$$( 2 x + 1 ) ^ { 2 } - x ^ { 2 } = 0$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=-1$$

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}.

$$4x^{2}+4x+1-x^{2}=0$$

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

$$3x^{2}+4x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\times 1=3$$

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.

$$a=1 b=3$$

Rewrite 3x^{2}+4x+1 as \left(3x^{2}+x\right)+\left(3x+1\right).

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

Factor out x in 3x^{2}+x.

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

Factor out common term 3x+1 by using distributive property.

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

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

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

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}.

$$4x^{2}+4x+1-x^{2}=0$$

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

$$3x^{2}+4x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 4 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3}}{2\times 3}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-4±\sqrt{16-12}}{2\times 3}$$

Add 16 to -12.

$$x=\frac{-4±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$x=\frac{-4±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-4±2}{6}$$

Now solve the equation x=\frac{-4±2}{6} when ± is plus. Add -4 to 2.

$$x=-\frac{2}{6}$$

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-4±2}{6} when ± is minus. Subtract 2 from -4.

$$x=-\frac{6}{6}$$

Divide -6 by 6.

$$x=-1$$

The equation is now solved.

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

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}.

$$4x^{2}+4x+1-x^{2}=0$$

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

$$3x^{2}+4x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Add -\frac{1}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{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}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x+\frac{2}{3}=\frac{1}{3} x+\frac{2}{3}=-\frac{1}{3}$$

Subtract \frac{2}{3} from both sides of the equation.

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