SOLUTION: Seriously stumped here, Please help me. I have to describe and correct the error. 2b^2+b=6 b^2+1/2b=3 b^2+1/2b+(1/2)^2=3+(1/2)^2 (b+1/2)^2=3+1/4 b+1/2=+sqrt13/4 b=-1/2+s

Algebra ->  Radicals -> SOLUTION: Seriously stumped here, Please help me. I have to describe and correct the error. 2b^2+b=6 b^2+1/2b=3 b^2+1/2b+(1/2)^2=3+(1/2)^2 (b+1/2)^2=3+1/4 b+1/2=+sqrt13/4 b=-1/2+s      Log On


   

Question 952900: Seriously stumped here, Please help me.
I have to describe and correct the error.
2b^2+b=6
b^2+1/2b=3
b^2+1/2b+(1/2)^2=3+(1/2)^2
(b+1/2)^2=3+1/4
b+1/2=+sqrt13/4
b=-1/2+sqrt13/2
b=-1+sqrt13dividedby 2 this is all the one problem..says to describe the error and correct it?
Found 2 solutions by Alan3354, macston:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
2b^2+b=6
b^2+1/2b=3
b^2+1/2b+(1/2)^2=3+(1/2)^2
(b+1/2)^2=3+1/4
b+1/2=+sqrt13/4
b=-1/2+sqrt13/2
b=-1+sqrt13dividedby 2 this is all the one problem..says to describe the error and correct it?
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(b + 1/2)^2 = 13/4
b + 1/2 = sqrt(13)/2
b+=+%28-1+%2B-sqrt%2813%29%29%2F2
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There's no error, but there are 2 solutions. + and minus

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
2b%5E2%2Bb=6
b%5E2%2B%281%2F2%29b=3
b%5E2%2B%281%2F2%29b%2B%281%2F2%29%5E2=3%2B%281%2F2%29%5E2 Should add %281%2F4%29%5E2 instead of %281%2F2%29%5E2 here.
b%5E2%2B%281%2F2%29b%2B%281%2F4%29%5E2=3%2B%281%2F4%29%5E2 This is what it should be.
%28b%2B1%2F4%29%5E2=3%2B1%2F16 GIVEN:%28b%2B1%2F2%29%5E2 would equal b%5E2%2B%28b%29%2B1%2F4 and NOT b%5E2%2B%281%2F2%29b%2B%281%2F4%29 The co=efficient of b would be 1, not 1/2.
b%2B1%2F4=sqrt%2849%2F16%29
b%2B1%2F4=7%2F4
b=6%2F4=1.5