Question

Solve the equation by completing square method.

$b^2+6b=41$

Short answer

The solutions are$b=-3+5\sqrt{2},b=-3-5\sqrt{2}$

Step-by-step solution

Given Information    

The Given equation is$b^2+6b=41$.

Isolate the variable terms on one side and the constant terms on the other. 

The variable terms are on the left side.

$b^2+6b=41$

Here, $b=6$.

Find 12b2.

$$\begin{gathered} ={\left(\frac{1}{2}\left(6\right)\right)}^2 \\ ={\left(3\right)}^2 \\ =9 \end{gathered}$$

Add $9$to both sides of equation.

$$\begin{gathered} b^2+6b+9=41+9 \\ {b}^2+6b+9=50 \end{gathered}$$

Factor the perfect square trinomial as a binomial square.    

$(b+3{)}^2=50$

Use the square root property.

$$\begin{gathered} b+3=\pm \sqrt{50} \\ b+3=\pm \sqrt{2\times 5\times 5} \\ b+3=\pm 5\sqrt{2} \end{gathered}$$

Solve for b.

$b=-3\pm 5\sqrt{2}$

Rewrite to show two solutions.

$$\begin{gathered} b=-3+5\sqrt{2}, \\ b=-3-5\sqrt{2} \end{gathered}$$