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. 2x^2+x<15

Accepted Answer

Hello today we're going to be solving the given polynomial inequality. So what we are given is three X squared plus 13 X. Less than 10. And whenever we want to solve this type of inequality, we want to go ahead and move all our terms to one side of the inequality. So in order to solve this inequality we're going to we're going to subtract 10 to both sides of the inequality and that's going to give us three X squared plus 13, X minus 10, less than zero. Now that we've moved all of our terms to one side. We want to go ahead and simplify the given polynomial three X squared plus 13 X minus 10 is a fact arable. Try no meal as this try no meal is going to factor to the quantities three x minus two times X plus five. And now that the polynomial is simplified. We need to go ahead and figure out the zeros for X. In order to plot them on the number line, in order to find the zeros for X. We're going to take each factor of X. Set equal to zero and solve for X. So doing this is going to give us three x minus two equal to zero and X plus five equal to zero on the left hand side, we're going to add two to both sides of the equation. And that's going to give us three, X is equal to two. And in order to solve for X, we're going to divide both sides by three and that's going to give us X is equal to 2/3, which is the first zero for X. And on the right hand side, we're just going to subtract five to both sides of the equation and that's going to give us X is equal to negative five, which is the next zero for X. Now that we have the zeros for X, we can plot them on a number line and use that to figure out the regions of the number line that satisfied the given inequality. So this number line is going to start a negative infinity and go to our first zero which is negative five. Then it's gonna go to negative five to our next zero, which is 2/3 And 2/3 is approximately equal to 0.6. And finally, it's going to go from 2/3 and end with positive infinity and notice that the original inequality is using the less than symbol, since we are using the less than symbol, we're not going to include the zeros in our solution set. And we're going to represent that with open circles on the number line. Now that we have the number line, we can go ahead and use this to figure out the regions that satisfied the given inequality. And in order to do that, we're going to take a testing point from each region, plug it into our original inequality and whichever one satisfies is going to be the regions that are included. So let's go ahead and test the region that's to the left of negative five. And the testing point from that region can be when X is equal to negative six. When X is equal to negative six, we get three times negative six squared plus times negative six should be less than 10. Negative six squared is going to give us positive 36 3 times 36 is going to give us 108. 13 times negative six is going to give us negative And 108 -78 is going to give us 30 less than 10, which is a false inequality. So since that testing point didn't satisfy the original inequality, we're not going to include any values that are to the left of negative five. Now, let's pick a testing point between negative five and 2/3. And that can be when X is equal to zero. When x is equal to zero, we get three times zero squared plus 13 times zero is less than 10. And on the left hand side, anything that's multiplied by zero is just going to zero out. So what we have is zero plus zero, which is just going to equal to zero. And the inequality we get is zero less than 10, which is true. So since the testing point between negative five and two or three satisfies the inequality, we're going to include the region between negative five and 2/3. Now let's pick a testing point to the right of 2/3 and this can be when x is equal to positive one. When X is equal to one, we get three times one squared plus 13 times one should be less than 10. One squared is just going to give us one and three times one is going to give us 3. 13 times one is going to give us 13 and three plus 13 is going to give us 16 less than 10, which is a false inequality. And since that's a false inequality, we're not going to include any values to the right of 2/3. Now that we figured out the regions that satisfied the given inequality, let's represent our answer in interval notation. So on the number line, there's only one region of values that satisfies the given inequality and that's gonna be the region that starts a negative five and ends at 2/3. So again, this region starts at -5. But since negative five is not being included in our solution set, we're going to assign it a parentheses. Then the region ends at positive 2/3. And since positive 2/3 is not being included, we assign it a parentheses. And this is going to be the solution to the given inequality. 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. 2x^2+x<15

Key Concepts

Polynomial Inequalities

Interval Notation

Graphing Solution Sets

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