Solve each polynomial inequality in Exercises 1–42 and graph the ...

Question

Solve each polynomial inequality in Exercises 1–42 and graph the solution set on a real number line. Express each solution set in interval notation. (2−x)^2(x−7/2)<0

Accepted Answer

Hello today we're going to be solving the given polynomial inequality. So what we are given is five minus X, quantity squared times x minus 13 over to less than zero. Now in order to solve this inequality I need to first make sure that the inequality is completely factored, which in this case is already given to us in a factored form. So what I'm going to need to do is figure out the regions on the number line that satisfy the inequality. But in order to create the number line, I'm going to need to first figure out the zeros for X. In order to get the zeros for X. I'm going to take each factor of the inequality set it equal to zero and solve for X. Doing this is going to give me five minus X quantity squared equal to zero and x minus 13/2 equal to zero. On the left equation. I can take the square root of both sides and that's going to give me five minus X equal to the square root of zero which is zero. Finally, I can add X to both sides of the equation. And this is going to give me X is equal to five, which is going to be my first zero. And on the second equation I can just add 13/2 to both sides. And this is going to give me X is equal to positive 13/2 and 13/2 can also be rewritten as 6. and that's gonna be my next zero. So now let's create the number line. So we're going to have a number line and going from left to right starts at negative infinity. Going to our first zero, which is going to be five, then from five to our next zero, which is going to be 13/2, which is also equal to 6.5. And then from this zero to positive infinity and notice that the inequality uses the less than sign. Since we're using the less than sign, we're not going to include the zeros in our solution set and we can represent that with open circles on the number line. So what we're going to want to do now is take a testing point from the three regions on the number line and plug them into our original inequality whichever one satisfies the inequality is going to be the regions that we include in our solution set. So let's go ahead and test this region that is to the left of five. And this is going to be when we have a testing point x is equal to zero, since zero is to the left of five. If we plug in zero into the inequality we get five minus zero squared times zero minus 13/2, which is also 6.5 and that's going to be less than zero, 5 0 is going to give me five And -6.5 is going to give me negative 6.5. Now five squared is going to give me 25 25 times negative 6.5 is going to give me negative 162.5 which is less than zero. So since that region satisfies the inequality, we're going to include the region that's to the left of five. Now let's go ahead and pick a testing point between five and 13/2. And that can be when X is equal to six. When X is equal to six, we get five minus six squared times six minus 13/2, which is also 6.5, less than zero five minus six is going to give me negative one and negative one squared is going to give me positive one and six minus 6.5 is going to give me negative 0.5 and negative 0.5 times one is going to give me negative 0.5 less than zero, which is true. So we're going to want to include the region of numbers that's between five and over to Finally, let's test the region that's to the right of 13/2 and we can let that be when X is equal to seven. When x is equal to seven we get five minus seven squared times seven minus 13/2, which is also 6.5, less than zero, 5 -7 is going to give me - And 7 -6.5 is going to give me 0. Native two squared is going to give me positive four and positive four times 0.5 is going to give me two less than zero, which is false because a positive number can never be less than zero. So since this region didn't satisfy the inequality, we're not going to include any values to the right of 13/2. Now let's go ahead and represent this answer in interval notation. So we're going to want to include the region going from negative infinity to positive five, but we're not going to include positive five. So the way we start that is by writing negative infinity. And since negative infinity doesn't have a certain value to it. We're going to assign it a parentheses. Then we're going to end this region with positive five. But since positive five is not being included, we're going to give it a parentheses. Then we're going to union the next set of the next interval. And that's going to be the region between five and 13/2, not including the end points. So again, that region starts with five and since five is not included, we give it a parentheses E. And we're going to end that region in 13/2. And since 13/2 is not included. We're going to close it with the parentheses E and this is going to be the solution of the inequality in interval notation. And with that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem, and I'll go ahead and see you all in the next video.

Solve each polynomial inequality in Exercises 1–42 and graph the solution set on a real number line. Express each solution set in interval notation. (2−x)^2(x−7/2)<0

Key Concepts

Polynomial Inequalities

Interval Notation

Graphing Solution Sets

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