A Quadratic InequalityDate: 4/27/96 at 17:10:42 From: Anonymous Subject: Quadratic Inequality Solve the inequality, and express the solutions in terms of intervals whenever possible. 1. (3x+1)(d-10x)>0 x > -(1/3) x < (1/2) I am able to solve for an answer but I do not understand how to express it in intervals. Date: 8/4/96 at 16:31:47 From: Doctor Mike Subject: Re: Quadratic Inequality The (d-10x) was probably intended to be (5-10x), right? That would match up with your partial answer x<(1/2). You are correct in the way that you have solved the problem so far. What you need to understand is that BOTH x > -(1/3) AND x < (1/2) has to be true at the same time for the same value of x. If you have such an x value, that x value will make (3x+1) > 0 and (5-10x) > 0 so that the product will be > 0 too. Where the interval part of the question comes into the picture is that any x value between -(1/3) and +(1/2) will work. The interval on the number line from -(1/3) to +(1/2) is usually represented as ( -(1/3), 1/2 ). Sometimes you see intervals represented like this: [ -(1/3), 1/2 ] using square brackets instead of parentheses. That means that the end-points, in this case the numbers -(1/3) and 1/2, are included in the interval. In your problem you DO NOT want to write the interval in this way because including these two x values will make the product equal to zero instead of greater than zero. If your original problem had been written " > = 0 " meaning greater-than-or- equal-to-zero, only then would the correct interval answer have been [ -(1/3) , 1/2 ]. Hope this helps. Please write back if you have other math questions and we will try to be faster in answering them. -Doctor Mike, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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