Question

Solve the equation by completing square method.$b^2=11b-5$

Short answer

The solutions are$b=\frac{-\sqrt{101}+11}{2},\frac{\sqrt{101}+11}{2}$

Step-by-step solution

Given Information.   

The Given equation is $b^2=11b-5$

Isolate the variable terms on one side and the constant terms on the other.The variable terms are on the left side.

$b^2-11b=-5$

The coefficient of $b=-11$

So ,$b=-11$

Find 12b2and add to the both sides of equation.

$$\begin{gathered} ={\left(\frac{1}{2}\left(-11\right)\right)}^2 \\ ={\left(\frac{-11}{2}\right)}^2 \\ =\frac{121}{4} \end{gathered}$$

Add $\frac{121}{4}$on both side of equation

$$\begin{gathered} b^2-11b+\frac{121}{4}=-5+\frac{121}{4} \\ {b}^2-11b+\frac{121}{4}=\frac{101}{4} \end{gathered}$$

Factor the perfect square trinomial as a binomial square  

${\left(b-\frac{11}{2}\right)}^2=\frac{101}{4}$

Use square root property

$$\begin{gathered} \left(b-\frac{11}{2}\right)=\pm \sqrt{\frac{101}{4}} \\ b=\frac{\pm \sqrt{101}}{2}+\frac{11}{2} \\ b=\frac{-\sqrt{101}+11}{2},\frac{\sqrt{101}+11}{2} \end{gathered}$$