Describe and correct the error`` 2b^2+b=6 b^2+(1/2)b=3 b^2+(1/2)b+(1/2)^2=3+(1/2)^2 (b+(1/2))^2=3+(1/4) b+(1/2)=`+-` `sqrt(13/4)` b=(-1/2) `+-` (`sqrt(13)` /2) b= (-1`+-` `sqrt(13)` /2)

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

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There is an error in the process of completing the square on the left hand side while going from line 3 to line 4, since

`b^2+1/2b+(1/2)^2!=(b+1/2)^2.` 

A correct solution starting at line 3 would be

`b^2+1/2b+(1/4)^2=3+(1/4)^2`

`(b+1/4)^2=3+1/16`

`b+1/4=+-sqrt(49/16)`

`b=-1/4+-7/4`.

This gives the two solutions `b=3/2` and `b=-2.`

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