Solve (2m+1)^2-(m^2-1)=1

$$( 2 m + 1 ) ^ { 2 } - ( m ^ { 2 } - 1 ) = 1$$

Solve for m

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

$$m=-1$$

Steps Using Factoring By Grouping

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

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

To find the opposite of m^{2}-1, find the opposite of each term.

$$4m^{2}+4m+1-m^{2}+1=1$$

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

$$3m^{2}+4m+1+1=1$$

Add 1 and 1 to get 2.

$$3m^{2}+4m+2=1$$

Subtract 1 from both sides.

$$3m^{2}+4m+2-1=0$$

Subtract 1 from 2 to get 1.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3m^{2}+am+bm+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 3m^{2}+4m+1 as \left(3m^{2}+m\right)+\left(3m+1\right).

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

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

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

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

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

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

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

Steps Using the Quadratic Formula

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

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

To find the opposite of m^{2}-1, find the opposite of each term.

$$4m^{2}+4m+1-m^{2}+1=1$$

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

$$3m^{2}+4m+1+1=1$$

Add 1 and 1 to get 2.

$$3m^{2}+4m+2=1$$

Subtract 1 from both sides.

$$3m^{2}+4m+2-1=0$$

Subtract 1 from 2 to get 1.

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

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}.

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

Square 4.

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

Multiply -4 times 3.

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

Add 16 to -12.

$$m=\frac{-4±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$m=\frac{-4±2}{2\times 3}$$

Multiply 2 times 3.

$$m=\frac{-4±2}{6}$$

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

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

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

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

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

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

Divide -6 by 6.

$$m=-1$$

The equation is now solved.

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

Steps for Completing the Square

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

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

To find the opposite of m^{2}-1, find the opposite of each term.

$$4m^{2}+4m+1-m^{2}+1=1$$

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

$$3m^{2}+4m+1+1=1$$

Add 1 and 1 to get 2.

$$3m^{2}+4m+2=1$$

Subtract 2 from both sides.

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

Subtract 2 from 1 to get -1.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

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.

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

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

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

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.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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