Solve 1+2k+k^2=0

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

Solve for k

$$k=-1$$

Steps Using Factoring

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor k^{2}+2k+1 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+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(k+a\right)\left(k+b\right) using the obtained values.

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

Rewrite as a binomial square.

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

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

$$k=-1$$

Steps Using Factoring By Grouping

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk+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 k^{2}+2k+1 as \left(k^{2}+k\right)+\left(k+1\right).

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

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

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

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

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

Rewrite as a binomial square.

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

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

$$k=-1$$

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.

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

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}.

$$k=\frac{-2±\sqrt{2^{2}-4}}{2}$$

Square 2.

$$k=\frac{-2±\sqrt{4-4}}{2}$$

Add 4 to -4.

$$k=\frac{-2±\sqrt{0}}{2}$$

Take the square root of 0.

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

Divide -2 by 2.

$$k=-1$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Factor k^{2}+2k+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$k+1=0 k+1=0$$

Subtract 1 from both sides of the equation.

$$k=-1 k=-1$$

The equation is now solved. Solutions are the same.

$$k=-1$$