Satyadeep A.

asked • 06/05/15

Find x such that 3x²-7x+6<0

This question belongs from rational inequalities.

2 Answers By Expert Tutors

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Michael J. answered • 06/06/15

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Effective High School STEM Tutor & CUNY Math Peer Leader

Andrew M.

Michael;
I keep going over this problem because I get a different y coordinate for the vertex when I use the complete the square method than when I simply plug in 7/6 for the x value.  As shown, plugging x = 7/6 into the original equation gives y = 23/12.... however when I complete the square to get format of y = a(x-h)2+k   I get y coordinate at vertex as 121/12. 
 
3x2-7x+6 = 3(x2-(7/3)x) +6 = 3(x2 - (7/3)x + 49/36) + 6 +3(49/36)
= 3(x-7/6)2 + 121/12
 
Thus it appears the vertex should be (7/6 , 121/12)  instead of  (7/6 , 23/12)
 
And I can find no error in either method...  How did you not add the 6 back in when you completed the square?
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06/06/15

Michael J.

Lets start from this part of my solution.
 
y = 3(x2 - (7/3)x + (49/36)) + (23/12)
 
The term in bold parenthesis is a perfect square.  Then I add a k value so that when I expand the whole thing out, my combined constants will be
 
3(49/36) + (23/12) = 6
 
6 is the constant term in the original function.
 
 
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06/06/15

Andrew M.

I appreciate it Michael.  I understood all of the problem I just made a rookie mistake when I added the 3(49/36) to complete the square I should have subtracted it on the outside instead of adding.
Andy
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06/08/15

Andrew M. answered • 06/05/15

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Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors

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