## What you need to know

Quadratic inequalities are precisely what they sound like: inequalities that involve a squared term, and therefore a quadratic. Much like normal quadratics, they usually have two-part solutions rather than just one. As with linear equalities, we can manipulate them to find solutions as if they were equations, but in the case of a quadratic inequality this gets a little more complicated. Also, depending on which sign is used in your inequality, your solutions can look quite different.

Before going further, you should be familiar both with linear inequalities (https://mathsmadeeasy.co.uk/gcse-maths-revision/inequalities-number-line-solving-inequalities/) and with using factorisation to solve quadratics (https://mathsmadeeasy.co.uk/gcse-maths-revision/solving-quadratics-factoring/). Now, let’s get into it.

Example: Solve the inequality $x^2-2x-3<0$.

The first step is to factorise this quadratic and find the solutions as if it were an equation, not an inequality. So, observing that $1\times(-3)=-3$ and $1+(-3)=-2$, we get that $x^2-3x-3=(x+1)(x-3)$ and so our inequality becomes

$(x+1)(x-3)<0$

So, if this were an equation, then the solutions to the quadratic would be $x=-1$ and $x=3$. Using this information, we can sketch the graph of $y= x^2-2x-3$ – this is the step that will help us understand how to solve the inequality.

On the right is a plot of the quadratic, and obviously I cheated by using a computer, but honestly it can be a rough a sketch as you want because what’s important is as follows. The question is asking for a solution to $x^2-2x-3<0$. In other words, when does the value of the quadratic go below zero? Now, the graph is a sketch of $y=x^2-2x-3$, so the question of when $x^2-2x-3$ goes below simply becomes: when does the graph go below zero on the $y$-axis? (In other words, when does the graph go below the $x$-axis?

Looking at the graph, we can see that when $x$ is bigger than -1 but smaller than 3, the graph is below zero (or, equivalently, below the $x$-axis). Therefore, the inequality is satisfied for values of $x$ that fall strictly in between -1 and 3, so we write the solution as

$-1.

Note: If the original inequality were $x^2-2x-3\leq0$ then the solution would’ve been $-1\leq x\leq3$. The inequality in the question corresponds to the one in the answer.

Now, let’s see what happens when the inequality is the opposite way around.

Example: Solve the inequality $x^2+5x\geq-4$. Write your answer in set notation.

Firstly, we need to rearrange this so it looks more like a familiar quadratic. Add 4 to both sides to get

$x^2+5x+4\geq0$

Now we can proceed with this process in the same way that we did for the last one. Observing that $1\times4=4$ and $1+4=5$, we can factorise this to be

$(x+1)(x+4)\geq0$

So, if this were an equation, then the solutions to the quadratic would be $x=-1$ and $x=-4$. Using this information, we can sketch the graph of $y=x^2=5x+4$.

The idea here is the same as the last example – we will use the graph as a reference to help us solve the inequality – but this time, the question is a little different. Since the inequality is now $x^2+5x+4\geq0$, the question becomes: when does the graph go above zero on the $y$-axis? (In other words, when does the graph go above the $x$-axis?)

We can see that when $x$ is less than -4, the graph is above zero, then it goes below zero, turns around, and reaches above zero again when it gets bigger than -1. So, we’re not look at one singular solution, we’re looking at 2. The inequality is satisfied when

$x\leq-4\,\text{ or }\,x\geq-1$.

This solution has a different structure to the last one, and you must include the ‘or’. It isn’t possible for $x$ to be both below -4 and above -1 at the same time, so the ‘or’ is required.

Now, to display the answer in set notation the process is the same as always: write an $x$ followed by a colon and the inequality, and then contain the whole thing inside curly brackets. Doing so, we get

${x: x\leq-4\,\text{ or }\,x\geq-1}$.

Note: at some point you may see a quadratic inequality with a negative $x^2$ term, in which case you can either do a) continue on as normal and remember that the graph will be an upside-down version of the ones above, or b) take everything over to the other side of the inequality and then the question becomes precisely the same as the ones seen above. Either way, don’t fret – it’s just a minor obstacle.

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So, we will factorise this quadratic, and then use what would be the solutions to help us plot the graph. Observing that $(-2)\times(-3)=6$ and $(-2)+(-3)=-5$, we get that

$x^2-5x+6=(x-2)(x-3)$,

and therefore, our inequality becomes

$(x-2)(x-3)\leq0$.

So, the “solutions” of this quadratic would be $x=2$ and $x=3$, therefore the graph looks like

The inequality in the question is $x^2-5x+6\leq0$, so the question is: when does the graph go below zero? We can see that it goes below zero when $x$ is between 2 and 3. Therefore, the solution to the inequality is

$2\leq x\leq3$.

Writing this in set notation, we get

$x:2\leq x\leq3$

So, we will factorise this quadratic, and then use what would be the solutions to help us plot the graph. Taking $x$ out as a factor, we get

$x^2-3x=x(x-3)$,

and therefore, our inequality becomes

$x(x-3)>0$.

So, the “solutions” of this quadratic would be $x=0$ and $x=3$, therefore the graph looks like

The inequality in the question is $x^2-3x>0$, so the question is: when does the graph go above zero? We can see that it goes above zero when $x$ is less than 0 and also when $x$ is above 3. Therefore, the solution to the inequality is

$x<0\,\text{ or }\,x>3$.

Level 8-9
Level 8-9