Solve (lambda+1)^2=0

$$( \lambda + 1 ) ^ { 2 } = 0$$

Solve for λ

$$\lambda =-1$$

Steps Using Factoring

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

$$\lambda ^{2}+2\lambda +1=0$$

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

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

Rewrite as a binomial square.

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

To find equation solution, solve \lambda +1=0.

$$\lambda =-1$$

Steps Using Factoring By Grouping

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

$$\lambda ^{2}+2\lambda +1=0$$

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

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

Factor out \lambda in \lambda ^{2}+\lambda .

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

Factor out common term \lambda +1 by using distributive property.

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

Rewrite as a binomial square.

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

To find equation solution, solve \lambda +1=0.

$$\lambda =-1$$

Steps Using the Quadratic Formula

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

$$\lambda ^{2}+2\lambda +1=0$$

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}.

$$\lambda =\frac{-2±\sqrt{2^{2}-4}}{2}$$

Square 2.

$$\lambda =\frac{-2±\sqrt{4-4}}{2}$$

Add 4 to -4.

$$\lambda =\frac{-2±\sqrt{0}}{2}$$

Take the square root of 0.

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

Divide -2 by 2.

$$\lambda =-1$$

Steps for Completing the Square

Take the square root of both sides of the equation.

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

Simplify.

$$\lambda +1=0 \lambda +1=0$$

Subtract 1 from both sides of the equation.

$$\lambda =-1 \lambda =-1$$

The equation is now solved. Solutions are the same.

$$\lambda =-1$$