Asked by miss b
Apparently, I got this question wrong on a test and I'd like to know what error I made... can you help?
Here's what I did:
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=+/- <13/4>
b=-1/2+/-<13>/2
b=(-1+/-<13>)/2
<> means its in a radical and +/- means plus or minus
Here's what I did:
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=+/- <13/4>
b=-1/2+/-<13>/2
b=(-1+/-<13>)/2
<> means its in a radical and +/- means plus or minus
Answers
Answered by
Reiny
The error is in your third line
you take 1/2 of the middle coefficient, and then square it
(1/2)(1/2) = 1/4 --- (1/4)^2 = 1/16
3rd line:
b^2 + (1/2)b + 1/16 = 3 + 1/16
(b+1/4)^2 = 49/16
b + 1/4 = ± 7/4
b = -1/4 ± 7/4 = 6/4 or -8/4
<b>b = -2 or 3/2</b>
since we have rational answers, we know it would have factored to
(x+2)(2b-3)
= 2b^2 + b - 6 , the original.
you take 1/2 of the middle coefficient, and then square it
(1/2)(1/2) = 1/4 --- (1/4)^2 = 1/16
3rd line:
b^2 + (1/2)b + 1/16 = 3 + 1/16
(b+1/4)^2 = 49/16
b + 1/4 = ± 7/4
b = -1/4 ± 7/4 = 6/4 or -8/4
<b>b = -2 or 3/2</b>
since we have rational answers, we know it would have factored to
(x+2)(2b-3)
= 2b^2 + b - 6 , the original.
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