Michael J. answered 06/06/15
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Effective High School STEM Tutor & CUNY Math Peer Leader
Suppose we graphed the function 3x2 - 7x + 6. According to the inequality, we want to find x values in which the y value is below the x-axis or are negative. If this was an equation, the values of x would have y values that are on the x-axis.
First, lets solve for the zeros by replacing the less than sign with an equal sign. We will use these zeros later on.
3x2 - 7x + 6 = 0
Use the quadratic formula to solve for the zeros.
x = (7 ± √(49 - 4(18))) / 6
x = (7 ± √(-23)) / 6
Note that we have a negative number under the square-root. This means that we will have a complex number. Complex numbers never touch the x-axis. Also, since we have a positive parabola, this graph will never cross the x-axis, meaning that there are no values of x in which the y values are below the x-axis.
Another way to show that the graph is never below the x-axis we put this graph in vertex form.
y = a(x - h)2 + k
where the vertex is (h, k).
y = 3(x2 - (7/3)x + (49/36)) + (23/12)
y = 3(x - (7/6))2 + (23/12)
h = 7/6
k = 23/12
The vertex is (7/6 , 23/12). This is the minimum of the parabola. This minimum is above the axis because the y value is positive. Therefore, the graph never reaches below the x-axis. Thus, there are no x values that make the inequality true.

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.
06/06/15