Solve lambda^2+4lambda+4=0

$$\lambda ^ { 2 } + 4 \lambda + 4 = 0$$

Solve for λ

$$\lambda =-2$$

Steps Using Factoring

To solve the equation, factor \lambda ^{2}+4\lambda +4 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.

$$a+b=4 ab=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

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

$$\left(\lambda +2\right)\left(\lambda +2\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve \lambda +2=0.

$$\lambda =-2$$

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 \lambda ^{2}+a\lambda +b\lambda +4. To find a and b, set up a system to be solved.

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

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite \lambda ^{2}+4\lambda +4 as \left(\lambda ^{2}+2\lambda \right)+\left(2\lambda +4\right).

$$\left(\lambda ^{2}+2\lambda \right)+\left(2\lambda +4\right)$$

Factor out \lambda in the first and 2 in the second group.

$$\lambda \left(\lambda +2\right)+2\left(\lambda +2\right)$$

Factor out common term \lambda +2 by using distributive property.

$$\left(\lambda +2\right)\left(\lambda +2\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve \lambda +2=0.

$$\lambda =-2$$

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.

$$\lambda ^{2}+4\lambda +4=0$$

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

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

Square 4.

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

Multiply -4 times 4.

$$\lambda =\frac{-4±\sqrt{16-16}}{2}$$

Add 16 to -16.

$$\lambda =\frac{-4±\sqrt{0}}{2}$$

Take the square root of 0.

$$\lambda =-\frac{4}{2}$$

Divide -4 by 2.

$$\lambda =-2$$

Steps for Completing the Square

Factor \lambda ^{2}+4\lambda +4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$\lambda +2=0 \lambda +2=0$$

Subtract 2 from both sides of the equation.

$$\lambda =-2 \lambda =-2$$

The equation is now solved. Solutions are the same.

$$\lambda =-2$$