Question Video: Solving Radical Inequalities in One Variable Involving Factorisation Mathematics • 10th Grade

Find algebraically the solution set of the inequality √(𝑥² + 8𝑥 + 16) ≥ 3.

05:09

Video Transcript

Find algebraically the solution set of the inequality the square root of 𝑥 squared plus eight 𝑥 plus 16 is greater than or equal to three.

So the first thing we want to do if we want to solve this inequality is remove the square root sign. The way we do this is by squaring both sides of our inequality. And when we do this, what we’re left with is 𝑥 squared plus eight 𝑥 plus 16 is greater than or equal to nine. That’s because if we square three, we get nine. And if we square a square root, then we’d just get the value that’s within that square root.

So now what we want to do is find critical values. And to do that, what we’re gonna do is we’re gonna solve a quadratic. But we want the quadratic equal to zero. So what we’re gonna do first of all is subtract nine from each side of our inequality. And when we do that, we get 𝑥 squared plus eight 𝑥 plus seven. And that’s because 16 minus nine is seven is greater than or equal to zero. Okay, great. So it’s in the form that we wanted. So now, what we’re gonna do is we’re gonna act as if it’s equal to zero. So we’re gonna take away the inequality. And we’re gonna do that cause we want to solve the quadratic to find our critical values.

So if I want to solve an inequality like 𝑥 squared plus eight 𝑥 plus seven is equal zero, then we do this in a number of ways. We could use quadratic formula, we could use completing the square. However, for this one, just because it’s the easiest, we’re going to factor. And to factor, what we need to do is find two values whose product is positive seven and whose sum is positive eight because that’s the coefficient of our 𝑥. Well, the two factors that we’re gonna use are one and seven. And that’s because one multiplied by seven is equal to seven. So our product gives us positive seven. And one add seven is equal to eight, positive eight.

Okay, so we know that these are our factors. So this means that we can now rewrite our quadratic using parentheses. So we have 𝑥 plus one multiplied by 𝑥 plus seven is equal to zero. So now, we need to find out what the values of 𝑥 are going to be. And to do that, what we need to do is set each of our parentheses equal to zero. And that’s because if we have a result of zero, it means that one of our parentheses has to be zero. And that’s because zero multiplied by anything gives us zero.

Well, if we start with 𝑥 plus one is equal to zero and we subtract one from each side of the equation, we get 𝑥 is equal to negative one. And similarly, if we do that to the other equation, we’ve got 𝑥 plus seven is equal to zero, we get 𝑥 is equal to negative seven. Okay, great. So we’ve now found our critical values: 𝑥 equal to negative one and 𝑥 is equal to negative seven.

So if we draw a quick sketch of our quadratic, we’d have this shape here. So we got a U-shaped parabola. And that’s because we have a positive 𝑥 squared. And then, we’ve got it crossing the 𝑥-axis at negative seven and negative one, which are our critical values. So what we could also have is our 𝑦-intercept. We don’t need it if this question was worth showing if we’re doing a sketch. And we know that would be seven. And that’s because if we had 𝑦 is equal to 𝑥 squared plus eight 𝑥 plus seven and 𝑥 was equal to zero, it has zero squared, which is zero plus eight multiplied by zero which is zero plus seven. And so that means that would just give us seven. Okay, great.

But how does this help us with our inequality? Well, if you take a look at our inequality, we’re interested where our quadratic is greater than or equal to zero. So therefore, we do not want this section that’s underneath the 𝑥-axis. We only want a section of our graph, which is above the 𝑥-axis. So how can we represent this? Well, we could write it using inequality signs because we could say that 𝑥 is less than or equal to negative seven. Because we want all the values that are less than negative seven or 𝑥 is greater than or equal to negative one cause we want everything to the right of negative one. So that’s what we’d write it using the inequality signs.

So we could also show this in set notation. And we do this because the question asked us to find algebraically the solution set of the inequality. So we do that with all the real numbers. So that’s our ℝ minus and that means except then we’ve got the values between negative seven and negative one. But we’ve got brackets here because it means that we don’t include negative seven and negative one because, as we’ve already said, they could be in the values because 𝑥 could be less than or equal to negative seven or 𝑥 could be greater than or equal to negative one.

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