Quadratic Inequalities Worksheets | Questions and Revision | MME

# Quadratic Inequalities Worksheets, Questions and Revision

Level 8 Level 9

## What you need to know

Quadratic inequalities are inequalities that involve a squared term, and therefore a quadratic. Much like normal quadratics, they usually have two-part solutions.

As with linear inequalities, 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 the following topics:

### Take Note:

At some point you may see a quadratic inequality with a negative $x^2$ term, in which case you can either:

1. Use the same methods outlined below and remember that the graph will be an upside-down version, ‘n’ shape.
2. Take everything over to the other side of the inequality and then the question becomes a positive inequality with the steps the same at outlined below.

### Example 1: Solving Quadratic Inequalities

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

The first step is to factorise this quadratic and find the solutions. 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 $y= x^2-2x-3$

Below is a plot of the quadratic. 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? In other words, when does the graph go below zero on the $y$-axis?

Looking at the graph, we can see that when $x$ is bigger than -1 but smaller than 3, the graph is below zero. 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.

### Example 2: Solving Quadratic Inequalities

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

Add 4 to both sides to get the quadratic inequality in the form we like

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

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 $y=x^2=5x+4$.

Since the inequality is now $x^2+5x+4\geq0$, the question becomes: when does the graph go above zero on the $y$-axis?

We can see that when $x$ is less than -4 and bigger than -1 it is above zero. Therefore the quadratic 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

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

### Example Questions

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 inequality can be written as,

\begin{aligned}x^2-5x+6&\leq0 \\(x-2)(x-3)&\leq0\end{aligned}

So, the roots 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 is,

$2\leq x\leq3$

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

\begin{aligned}x^2-3x&>0 \\x(x-3)&>0\end{aligned}

So, the roots 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 is,

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

We solve this inequality by simply rearranging it to make $p$ the subject,

\begin{aligned}3p^2+8 &> 20 \\ 3p^2 &> 12 \\ p^2&>4 \end{aligned}

Hence $p$ can take any value greater than 2, $p>2$ or less than -2, $p<-2$, therefore the graph looks like,

The inequality in the question is $3p^2-12>0$, so the question is: when does the graph go above zero? We can see that it goes above zero when $p$ is less than -2 and also when $p$ is above 2. Therefore, the solution is,

$p<-2\,\text{ or }\,p>2$

We solve this inequality by simply rearranging it to make $x$ the subject,

\begin{aligned}x^2-3x+2&<0 \\ (x-2)(x-1)&<0 \end{aligned}

Hence $x$ can take any value between $1

The inequality in the question is $x^2-3x+2<0$, so the question is: when does the graph go below zero? We can see that it goes below zero between the values $x$ is greater than 1 and $x$ is less than 2. Therefore, the solution is,

$1

We solve this inequality by simply rearranging it to make $x$ the subject,

\begin{aligned}x^2-5x+24&<5x+8 \\ x^2-10x+16&<0 \\ (x-8)(x-2)&<0\end{aligned}

Hence $x$ can take any value between $2

The inequality in the question is $x^2-10x+16<0$, so the question is: when does the graph go below zero? We can see that it goes below zero between the values $x$ is greater than 2 and $x$ is less than 8. Therefore, the solution is,

$2

Level 8-9

Level 8-9