# Lesson video

In progress...

Hi, my name's Mr. Clasper.

And today we're going to learn how to solve a quadratic equation whose leading coefficient is greater than one.

Let's solve this equation.

So we know that the product of these two brackets must be equal to zero.

Therefore that means that either the value of the first bracket is equal to zero or the value of the second bracket is equal to zero.

So let's solve each of these separately.

We're going to solve two y plus seven is equal to zero and we're also going to solve y subtract two is equal to zero.

So starting with the equation on the left, two y plus seven is equal to zero.

So if we subtract seven from both sides, that must mean that two y is equal to negative seven.

If we divide both sides of this equation by two, this means that y must be equal to negative seven over two.

This is one of our solutions.

Looking at the equation on the right-hand side, we have y subtract two is equal to zero.

If we add two to both sides, this means that y must be equal to two.

This is our other solution.

So when we solve this equation, we get the solutions y is equal to negative seven over two and y is equal to two.

Let's solve this equation.

The first thing we need to do is to factorised it, so that we can solve it in a similar manner to the previous example.

Let's use a multiplication grid to help.

So we have two y and y.

When we multiply these, we get two y squared.

And we need two numbers which multiply to give us negative 18, but we need the same two numbers to generate a sum of negative nine y.

This combination will be three and negative six as this gives us three y and negative 12 y which have a sum of negative nine y.

So this means that our original expression on the left-hand side of our equation is equal to two y plus three multiplied by y minus six.

This means that we can solve this equation in this form.

So like we did on the previous example, we can solve two y plus three is equal to zero, then we can solve y minus six is equal to zero.

So starting with the equation on the left, two y plus three is equal to zero.

When we subtract three from both sides, we get two y is equal to negative three.

Then dividing both sides by two would mean that we have a solution of y is equal to negative three over two.

Looking at the equation on the right-hand side, we have y subtract six is equal to zero.

And if we add six to both sides, we get y is equal to six.

So our two solutions for this equation would be y is equal to negative three over two and y is equal to six.

Here are some questions for you to try.

Pause the video to complete your task and resume once you're finished.

Let's take a look at question two d.

So when we look at two d, we can see that there's only one solution which is x is equal to negative three over two.

This is because when we factorised the expression on the left-hand side of the equation, we would get two x plus three, or multiplied by two x plus three.

So because the brackets are the same, this generates the same solution.

And if we look at part e, we can see that our solutions are x is equal to five over two and x is equal to negative five over two.

This is because this is a difference of two squares.

So when we factorised this expression, we would get two x plus five and two x minus five.

Here's another question for you to try.

Pause the video to complete your task and resume once you're finished.

So we can see the bottom right card was the correct choice.

If we look at the top left, it's very similar.

However, we have solutions of positive three over five and negative two and they should be negative three over five and positive two.

If we look at the card to the top right, when we expand the two brackets given, we actually get five a squared plus 13 a plus six, which is not equivalent to the expression that we have on the left-hand side of the original equation.

And if I expand the brackets in the bottom left, we get five a squared minus 13 a plus six.

And again, this isn't equivalent to the expression on the left-hand side of our original equation.

And that brings us to the end of our lesson.

So I hope you're more confident with solving quadratic equations.

I will hopefully see you soon.