Video: Pack 5 β€’ Paper 1 β€’ Question 20

Pack 5 β€’ Paper 1 β€’ Question 20

07:08

Video Transcript

Solve the inequality π‘₯ squared is greater than or equal to two multiplied by four minus π‘₯.

This is a quadratic inequality, as the highest power of π‘₯ is two. There are a couple of specific methods that we need to follow when solving a quadratic inequality, which we’ll discuss in a moment.

First, we’ll rearrange this inequality slightly. We’ll begin by expanding the bracket on the right, giving π‘₯ squared is greater than or equal to eight minus two π‘₯. Next, we’ll group all the terms on the same side of the inequality by first adding two π‘₯ to both sides and then subtracting eight from both sides, to give π‘₯ squared plus two π‘₯ minus eight is greater than or equal to zero.

Now we’re going to look at two methods for solving a quadratic inequality. The first of these is to sketch the graph of 𝑦 equals π‘₯ squared plus two π‘₯ minus eight. This is a quadratic curve. And as the coefficient of π‘₯ squared is positive, we know that its shape will be a U-shaped parabola. We’ll think about the points where this graph intersects with the π‘₯- and 𝑦-axes in a moment.

The reason we sketch this graph is because we can then look at the regions where the graph is above or below the π‘₯-axis to help us with solving the inequality. Another way of describing the π‘₯-axis is that it’s the line 𝑦 is equal to zero. And 𝑦 is greater than zero everywhere above the π‘₯-axis and less than zero everywhere below the π‘₯-axis.

As we’ve written 𝑦 equals π‘₯ squared plus two π‘₯ minus eight, this also means that π‘₯ squared plus two π‘₯ minus eight is greater than zero above the π‘₯-axis and less than zero below the π‘₯-axis. That’s the sections of the graph that I’ve shaded in pink. The π‘₯ values that correspond to 𝑦 values which are greater than or equal to zero will be the sections of the π‘₯-axis that I’ve highlighted in orange.

In order to find the inequalities that describe these regions, we need to find the π‘₯ values of the coordinates of the points where the graph crosses the π‘₯-axis. To do so, we solve the quadratic equation π‘₯ squared plus two π‘₯ minus eight is equal to zero. This quadratic can be solved by factorising. As the coefficient of π‘₯ squared is one, we just have an π‘₯ as the first term in each of our two factorised brackets.

To complete the brackets, we’re looking for two numbers that sum to positive two and multiply to negative eight. Those numbers are positive four and negative two. So our quadratic factorises as π‘₯ plus four multiplied by π‘₯ minus two. To solve for π‘₯, we set each of these brackets in turn equal to zero. So we had π‘₯ plus four is equal to zero or π‘₯ minus two is equal to zero.

These two equations are straightforward to solve. We solve the first by subtracting four from both sides and solve the second by adding two to both sides, giving π‘₯ is equal to negative four and π‘₯ is equal to two. These values are sometimes called the critical values of the function. And they give the π‘₯-coordinates of the points where the graph crosses the π‘₯-axis. Note that we can also find the 𝑦-intercept of the curve by substituting π‘₯ equals zero into its equation, which gives 𝑦 is equal to negative eight.

Now that we found the π‘₯-coordinates of the points where the curve crosses the π‘₯-axis, we can find the inequalities that describe the two regions we’re looking for.

The first region on the left of the graph is all of the π‘₯ values from negative four downwards, so we can describe this as π‘₯ is less than or equal to negative four. The second region on the right of the graph is all of the π‘₯ values from two upwards, so we can describe this region as π‘₯ is greater than or equal to two. These two regions are entirely separate, so they can’t be combined into a single inequality. They give the solution to the original quadratic inequality: π‘₯ is less than or equal to negative four and π‘₯ is greater than or equal to two.

Now I mentioned that there was a second method that we can use when solving a quadratic inequality. It is to consider the sign of each of the individual brackets and then the sign of the overall function outside and in between the critical values. The critical values each make one of the individual brackets equal to zero. And therefore, they make the overall function equal to zero.

Let’s consider a value of π‘₯ less than negative four. So, for example, we could choose negative five. π‘₯ plus four and π‘₯ minus two will both be negative for this π‘₯ value. The product of two negative numbers is a positive number, so the whole function will be positive for π‘₯ values in this region.

Now let’s consider an π‘₯ value between the critical values. So, for example, we could choose an π‘₯ value of zero. π‘₯ plus four will be positive here, and π‘₯ minus two will be negative. The product of a positive and negative number is a negative number, so the whole function is negative for π‘₯ values between the two critical values.

Finally, we consider an π‘₯ value greater than the critical value of two. So, for example, we could use three. π‘₯ plus four and π‘₯ minus two will both be positive here. And the product of two positive numbers gives a positive, so the whole function will be positive here. This shows again that the whole function is positive or zero for all π‘₯ values less than or equal to negative four and π‘₯ values greater than or equal to two, which is the same as the solution we found by sketching the graph. You can use either of these two methods, but you must either sketch the quadratic graph or use a table of values.

Now there’s a common mistake that’s often made when solving quadratic inequalities. Remember the quadratic factorised as π‘₯ plus four multiplied by π‘₯ minus two. The common mistake is to try to solve a quadratic inequality in the same way as we solve a quadratic equation, so that’s solving each bracket individually being greater than or equal to zero. This would give two linear inequalities, the first of which can be solved by subtracting four from both sides and the second can be solved by adding two to both sides. This will give a solution of π‘₯ is greater than or equal to negative four and π‘₯ is greater than or equal to two.

However, notice that the direction of the inequality sign in our first inequality is opposite to what it should be. We have π‘₯ is greater than or equal to negative four, but the solution we’ve already found is π‘₯ is less than or equal to negative four. This is not a valid method for solving a quadratic inequality as, as we’ve seen, it leads to an incorrect answer. However, it is a very common mistake, so make sure that you don’t make it.

The correct solution to the inequality π‘₯ squared is greater than or equal to two multiplied by four minus π‘₯ is π‘₯ is less than or equal to negative four and π‘₯ is greater than or equal to two.

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