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# Find the set of values of x for which (3x-1)(x+3)<0?

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1. I know how you would do this if it was a =, but with > and < how do you work out which way round the sign goes in the final answer?
2. (Original post by Sint)
I know how you would do this if it was a =, but with > and < how do you work out which way round the sign goes in the final answer?
I would do a quick sketch of the graph, so when it equal 0. Then from you graph you can see which way the sign should be.
3. (Original post by Sint)
I know how you would do this if it was a =, but with > and < how do you work out which way round the sign goes in the final answer?
Find the critical values, by letting (3x-1)(x+3) = 0, find the values of x, x=1/3 or x=-3 and draw the graph. when it's > , you're looking for all values of x above the x axis, when it's < , it's all the points below the x axis.
4. You look at it graphically. You would work out the solutions which would be the same if it was = 0. Then look at the graph, the parts where the graph is above or below the x axis (depending on the > or < sign)

Hope this helps. Use desmos if you are struggling to sketch the graph.
5. (Original post by Sint)
I know how you would do this if it was a =, but with > and < how do you work out which way round the sign goes in the final answer?
Sketch the curve using the critical values for =0 as the points it crosses the x-axis
If a and b are those critical values then
You see clearly for <0 the values lie between the critical values a<x<b
For >0 you can see it lies outside the critical values so x<a or x>b

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6. First solve the equations (3x-1)(x+3)=0
(So basically solve 3x-1=0 and x+3=0)
You then get the roots and as this is a positive quadratic with the U shape you can know that the quadratic will be smaller than 0 between the roots so you'll get something like X1<x<X2 where X1 and X2 are the roots.
Hope it helped!
7. Cheers everyone, I know how to do it now!
8. Method for the pros: notice that the two coefficients of x in the brackets are the same sign, so the a coefficient of the quadratic is positive and the parabola is "U" shaped (rather than an upside down U when a is negative), so the part between the roots must be the negative bit (as the parabola will go to +inf as x --> +/- inf.

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Updated: October 9, 2016
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