Find all solutions to the inequality 𝑥 plus four squared is less than 136 minus nine 𝑥 plus four. Write your answer as an interval.
Let’s have a look at this inequality more closely. We can see that it is in fact a quadratic inequality because we have 𝑥 plus four all squared on the left-hand side. And when we distribute the parentheses, the highest power of 𝑥 that appears in this inequality will be two.
Now it would be possible to distribute the parentheses and then collect all terms on the same side of the inequality. But in fact there is a slightly easier away. We can see that 𝑥 plus four appears in two places in our inequality. We have 𝑥 plus four all squared on the left-hand side. And then we have negative nine multiplied by 𝑥 plus four on the right-hand side.
What we’re going to do is make a linear substitution. We’re going to introduce the variable 𝑢, which is equal to 𝑥 plus four. Our inequality then becomes slightly more straightforward. It’s still a quadratic inequality. But we now have 𝑢 squared is less than 136 minus nine 𝑢. Collecting all of the terms on the left-hand side, and we now have a quadratic inequality in its most easily recognizable form. 𝑢 squared plus nine 𝑢 minus 136 is less than zero.
Let’s see how we can solve this quadratic inequality. And we begin by seeing whether it’s possible to factor. As the coefficient of 𝑢 squared is one, the first term in each bracket will be 𝑢. And we’re then looking for two numbers which sum to the coefficient of 𝑢 — that’s nine — and whose product is the constant term — that’s negative 136.
We start by listing all the factor pairs of 136. We have one and 136, two and 68, four and 34, and eight and 17. And we then see that if we make the eight negative but keep the 17 positive, then the sum of these two numbers is indeed equal to nine. So our quadratic can be factored as 𝑢 plus 17 multiplied by 𝑢 minus eight. And this can of course be confirmed by distributing the parentheses again.
Next, we recall that, to solve a quadratic inequality in factored form, we need to consider a sketch of its graph. First, we find the critical values, which are also the 𝑥-intercepts of this quadratic graph. We find these by setting each of our factors in turn equal to zero. Giving 𝑢 plus 17 equals zero, which leads to 𝑢 equals negative 17, and 𝑢 minus eight equals zero, which leads to 𝑢 equals eight. So we find the critical values of this quadratic inequality are negative 17 and eight.
As we have a positive leading coefficient, that is, the coefficient of 𝑢 squared is positive one, we also know that the shape of this quadratic graph will be a positive parabola. Its 𝑦-intercept will be the value of the constant term — that’s negative 136. So combining all of these key features, we can produce a sketch of this quadratic, with critical values at negative 17 and eight and a negative 𝑦-intercept.
Now remember, we’re looking to solve where this quadratic is less than zero, which means we’re looking for where its graph lies below the 𝑢-axis. Highlighted in pink, we see that the graph lies below the 𝑢-axis for values of 𝑢 between the critical values of negative 17 and eight. So we can say that 𝑢 squared plus nine 𝑢 minus 136 is less than zero when 𝑢 is greater than negative 17 but less than eight.
Remember though that our original inequality wasn’t in terms of 𝑢. It was in terms of 𝑥. So we must make sure that we give our answer in terms of the original variable. We defined 𝑢 to take the place of the expression 𝑥 plus four. So replacing 𝑢 with 𝑥 plus four, we now have a solution in terms of 𝑥. Negative 17 is less than 𝑥 plus four, which is less than eight. Subtracting four from each part of our inequality, and we find that 𝑥 is greater than negative 21 but less than four. We can express this as the open interval from negative 21 to four.
So by first introducing a new variable 𝑢 to take the place of 𝑥 plus four and then finding the critical values of this quadratic so that we can sketch its graph. We’ve found that the solution to the given inequality is the open interval negative 21 to four. Remember, it would also have been possible to answer this question by distributing the parentheses and not replacing the variable 𝑥 plus four. Although this would’ve required some additional algebra and may have led to a more complicated quadratic to solve.