SOLUTION: Use completing the square to solve each equation. b^2 - 3b = 5 My answer is no where even close to what the book says it should be. I am in desperate need of help with these ty

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Use completing the square to solve each equation. b^2 - 3b = 5 My answer is no where even close to what the book says it should be. I am in desperate need of help with these ty      Log On


   

Question 44712This question is from textbook Intermediate Algebra
: Use completing the square to solve each equation.
b^2 - 3b = 5
My answer is no where even close to what the book says it should be. I am in desperate need of help with these types of problems. My book says the answer should be (3 + sqrt(29))/2 or (3 - sqrt(29))/2
This is what I came up with. What am I doing wrong?
b^2 - 3b = 5
+b%5E2+-+3b+%2B+%281%2F2%2A3%2F1%29%5E2+=+5+%2B+%281%2F2%2A3%2F1%29%5E2+
+b%5E2+-+3b+%2B+%283%2F2%29%5E2+=+5+%2B+%283%2F2%29%5E2+
+%28b+%2B+3%2F2%29%5E2+=+5+%2B+%283%2F2%29%5E2+
+%28b+%2B+3%2F2%29%5E2+=+5+%2B+%289%2F4%29+
+%28b+%2B+3%2F2%29%5E2+=+%285%2F4%29+%2B+%289%2F4%29+
+%28b+%2B+3%2F2%29%5E2+=+%2814%2F4%29+
+b+%2B+3%2F2+=%2Bplus_minus-%287%2F2%29+
+b+=+-%283%2F2%29%2Bplus_minus-%287%2F2%29+
 This question is from textbook Intermediate Algebra

Answer by adamchapman(301) About Me  (Show Source):
You can put this solution on YOUR website!
The first step should be to make the right hand side equal to zero:
b%5E2+-+3b+=+5
b%5E2-3b-5=0
Now try completing the square:
%28b-%283%2F2%29%29%5E2-5-%283%2F2%29%5E2=0
%28b-%283%2F2%29%29%5E2-5-%283%2F2%29%5E2=0
%28b-%283%2F2%29%29%5E2-5-%289%2F4%29=0
%28b-%283%2F2%29%29%5E2-20%2F4-9%2F4=0
%28b-%283%2F2%29%29%5E2-29%2F4=0
Now rearrange to find b:
%28b-3%2F2%29%5E2=29%2F4
b=3%2F2%2B-sqrt%2829%2F4%29
b=3%2F2%2B-sqrt%2829%29%2F2
b=%283%2B-sqrt%2829%29%29%2F2
I hope this helps.
P.S. I am trying to start up my own homework help website. I would be extremely grateful if you would e-mail me some feedback on the help you received to adam.chapman@student.manchester.ac.uk
Adam