Solve r^2+2r+1=0

$$r ^ { 2 } + 2 r + 1 = 0$$

Solve for r

$$r=-1$$

Steps Using Factoring

To solve the equation, factor r^{2}+2r+1 using formula r^{2}+\left(a+b\right)r+ab=\left(r+a\right)\left(r+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=1$$

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.

$$a=1 b=1$$

Rewrite factored expression \left(r+a\right)\left(r+b\right) using the obtained values.

$$\left(r+1\right)\left(r+1\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve r+1=0.

$$r=-1$$

Steps Using Factoring By Grouping

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

$$a+b=2 ab=1\times 1=1$$

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.

$$a=1 b=1$$

Rewrite r^{2}+2r+1 as \left(r^{2}+r\right)+\left(r+1\right).

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

Factor out r in r^{2}+r.

$$r\left(r+1\right)+r+1$$

Factor out common term r+1 by using distributive property.

$$\left(r+1\right)\left(r+1\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve r+1=0.

$$r=-1$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$r^{2}+2r+1=0$$

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

$$r=\frac{-2±\sqrt{2^{2}-4}}{2}$$

Square 2.

$$r=\frac{-2±\sqrt{4-4}}{2}$$

Add 4 to -4.

$$r=\frac{-2±\sqrt{0}}{2}$$

Take the square root of 0.

$$r=-\frac{2}{2}$$

Divide -2 by 2.

$$r=-1$$

Steps for Completing the Square

Factor r^{2}+2r+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}.

$$\left(r+1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+1\right)^{2}}=\sqrt{0}$$

Simplify.

$$r+1=0 r+1=0$$

Subtract 1 from both sides of the equation.

$$r=-1 r=-1$$

The equation is now solved. Solutions are the same.

$$r=-1$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +2x +1 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -2 rs = 1$$

Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1

$$(-1 - u) (-1 + u) = 1$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$1 - u^2 = 1$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = 1-1 = 0$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 0 u = 0$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r = s = -1$$