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Video: Solving an Inequality of Second Degree in One Variable in R

Alex Cutbill

Find all solutions to the inequality 𝑥² + 121 ≀ 0. Write your answer as an interval.

02:55

Video Transcript

Find all solutions to the inequality π‘₯ squared plus 121 is less than or equal to zero. Write your answer as an interval.

We have a quadratic inequality, and our first step therefore is to graph the quadratic we have, π‘₯ squared plus 121. What is the 𝑦-intercept? Well it’s a constant term, 121. So we can mark this on the graph.

Okay, how about the π‘₯-intercepts then? Finding those means solving π‘₯ squared plus 121 equals zero. We can apply the quadratic formula which tells us that the solutions to π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero are minus 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ all over two π‘Ž. In our case, π‘Ž is one, 𝑏 is zero. There’s no π‘₯ term there, and 𝑐 is 121. We can substitute those values in and simplify somewhat to get plus or minus the square root of negative 484 over two.

Here we’ve got the square root of a negative number, and we know that can’t possibly happen for real numbers. And so there are no real number solutions for our equation. This is because the discriminant, which is 𝑏 squared minus four π‘Žπ‘, is less than zero. We’ve discovered that the equation that was supposed to give us our π‘₯-intercepts has no solutions, and so we have to conclude that there aren’t any π‘₯-intercepts.

So what does the graph look like? The coefficient of π‘₯ squared is one which is positive, so we have an upward facing curve. And you can see that it makes sense that there are no π‘₯-intercepts, because the curve is going in the wrong direction to cross the π‘₯-axis. The graph of 𝑓 of π‘₯ equals π‘₯ squared plus 121 is just the graph of 𝑓 of π‘₯ equals π‘₯ squared, translate it up by 121 units. And so it makes sense that it doesn’t cross the π‘₯-axis.

So going back to the inequality that we’re trying to solve, π‘₯ squared plus 121 is less than or equal to zero. We can see from the graph that π‘₯ squared plus 121 is always greater than zero. The curve lies in the first two quadrants and doesn’t touch or cross the π‘₯-axis. There are, therefore, no solutions to this inequality. How can we express the idea of there being no solutions as an interval? We use this symbol here, which stands for the empty set. Interval notation is a special kind of set notation. And in fact, this empty set is indeed an interval technically, and it is our answer.