# Lesson video

In progress...

Hello, my name is Miss Parhnam, and in this lesson we're going to solve quadratic inequalities where a equals one.

So when we're asked to solve quadratic inequalities where a equals one we means that the x squared coefficient is one.

So for example, x squared plus two x, subtract three, greater than zero.

If this was a quadratic equation we would factorise that expression on the left.

And that's just what we'll do now.

So we can see we have a sum of two, that's the x coefficient, and a product of negative three, that's the constant value, which leads us to x add three, multiplied by x subtract one, all greater than zero.

So we can think about the roots of the graph on a quick sketch graph.

And we can see the two crossing points on the x-axis are zero, negative three, and zero, one.

A really rough curved sketched through those points.

So this is two distinct parts of the parabola.

X is either less than negative three or x is greater than one.

And we can represent this on a number line.

So we have a circle over one, unshaded because it does not include one, and the arrow points to everything greater than one.

And equally a circle over negative three and the arrow points to everything less than negative three.

And when we use set notation, because we have two distinct inequalities here that are not joined together we use the union symbol in between.

Here's a question for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers! We have two numbers with a sum of one and a product of negative six, so that naturally leads us to positive three and negative two into our brackets.

And if this has a product greater than zero than x must be less than negative three or greater than two.

So each set is distinct from the other.

So the diagram shows these separate sets and we use the union symbol in our set notation.

Here are some questions for you to try.

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So once you've factorised those quadratics, for part A, you should have x add two multiplied by x subtract two, either way around, multiplication is commutative.

And for part B, x add five multiplied by x subtract three.

Again, either way around.

And for part C, we have x add four multiplied by x subtract one.

Once again, it can be either way around.

And once you've factorised those that will help you get to the solutions and therefore we see diagrams and matching set notation.

So notice when it is one complete set we have a single line on the diagram with circles at each end and a double inequality in set notation, whereas if there's two distinct sets then we see two arrows with circles on the end of each and we use the union symbol in set notation.

Here's another question for you to try.

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Amir has a quadratic which is less than zero, so this is the section of the parabola underneath the x-axis, so this is one set of numbers.

On a number line this would be a single line with circles on each end.

And using set notation it would include a double inequality.

Here are some further questions for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers! In question five that curve is four above the x-axis at its turning point.

So no part of it is below the x-axis, hence no solutions.

That's all for this lesson.

Thank you for watching.