# Video: Solving Quadratic Inequalities in One Variable

Write the interval describing all solutions to the inequality 30 + 13𝑥 − 𝑥² > 0.

02:39

### Video Transcript

Write the interval describing all solutions to the inequality 30 plus 13𝑥 minus 𝑥 squared is greater than zero.

We have a quadratic inequality. And the first thing that we should do when trying to solve a quadratic inequality is to begin by solving a quadratic equation. Here, we’re going to set our quadratic expression equal to zero and solve for 𝑥. So we obtain 30 plus 13𝑥 minus 𝑥 squared equals zero. Usually, we would look to factor this expression. But the negative coefficient of 𝑥 squared makes that tricky. So instead, we’ll begin by multiplying through by negative one. And our equation becomes 𝑥 squared minus 13𝑥 minus 30 equals zero.

We’re now ready to factor the expression 𝑥 squared minus 13𝑥 minus 30. It’s a quadratic with no common factors aside from one. So we know it’s going to factor into two brackets. We know that the first term in each bracket will be 𝑥. And to find the second part of each binomial, we will need two numbers whose product is negative 30 and whose sum is negative 13. These are negative 15 and two. So we find our quadratic expression is equal to 𝑥 minus 15 times 𝑥 plus two.

Now, we know that the product of our expressions is equal to zero. And the only way for the product of two numbers to be equal to zero is if one number itself is zero. So we can say that either 𝑥 minus 15 is equal to zero. Or 𝑥 plus two is equal to zero. We would solve this first equation by adding 15 to both sides. And we find that 𝑥 is equal to 15. Similarly, we solve the second equation by subtracting two from both sides. And we find that 𝑥 is equal to negative two. And we have the two solutions to the equation 30 plus 13𝑥 minus 𝑥 squared equals zero.

But how does this help us solve the quadratic inequality? Well, one way is to sketch the graph of 𝑦 equals 30 plus 13𝑥 minus 𝑥 squared. We have found its roots to be 𝑥 equals 15 and 𝑥 equals negative two. These are the places where our graph crosses the 𝑥-axis. Similarly, it crosses the 𝑦-axis at 30. And since the coefficient of 𝑥 squared is negative, we have a sort of inverted parabola. And so our graph might look a little something like this.

Now, we’re trying to figure out where this graph is greater than zero. Let’s highlight the part of the graph that sits above the 𝑥-axis. Each value on this part of the curve is positive. Since we’re only interested in the values that are greater than zero, we’re not going to include the end points. That’s where 𝑥 is equal to negative two and 𝑥 is equal to 15. And so we see that the values of 𝑥 that make the expression 30 plus 13𝑥 minus 𝑥 squared greater than zero lie between negative two and 15.

In interval notation, we say that that’s the open interval from negative two to 15.