Lesson video

Hi, and welcome to today's lesson on solving equations with surds.

By the end of today's lesson you'll be able to solve an equation where some of the coefficients or terms are surds.

Our lesson has two parts and we're going to start with part one, which is solving linear equations.

Some equations may have surds as coefficients.

For example, this equation, 3x equals the square root of 5 multiplied by x plus 3.

Some equations may have surd solutions if we need the answer to be exact.

For example, here we can see x equals 1/4 multiplied by 9 root 5 plus 15.

We're going to find the solution to the equation 3x equals root 5 multiplied by x plus 3.

We're going to find our solution in the form a plus b root 5, where both a and b are rational numbers.

Now, when solving this sort of linear equation we know there are two approaches.

We can either divide by root 5 or we could expand the brackets.

It's entirely up to you which method you choose to go for.

Personally, I'm going to choose to expand the brackets here and that's because I know otherwise I'm going to end up with 3 over root 5 and I'm going to be ending up having to take away an x term.

I don't particularly want to do this, so instead, I'm going to expand the brackets as I suspect this will be an easier method to do.

But by all means pause the video and try the other way.

By expanding the brackets, I reach the following step.

3x is equal to root 5x plus 3 root 5.

Now, I know that my next step is to ensure that I can gather the like terms, and I can see there are two terms involving x, so let's go about gathering them.

First, I take away root 5x from both sides of the equation and I have 3x subtract root 5x equals 3 root 5.

I'm then going to factorise by taking out a factor of x.

This leaves me with x multiplied by 3 subtract root 5, and of course this is equal to 3 root 5.

Now to make x the subject of my equation, I need to divide.

So x is equal to 3 root 5 all over 3 subtract root 5.

Now of course there is an issue here in that although I have made x the subject of my equation, I don't have my solution in the form a plus b root 5.

Well, luckily I know what to do here.

I need to rationalise that denominator.

So let's do this now.

What will I need to multiply by? I'll need to multiply by 3 plus root 5 over 3 plus root 5.

Well, let's expand those brackets then.

My numerator will become 9 root 5 plus 15, and my denominator becomes 9 subtract 5.

Well that means my denominator is just equal to 4.

Now that's still not of the correct form, so I'll need to divide each term of the numerator by the denominator giving me 9 over 4 root 5 plus 15 over 4.

This is now of the correct form, so I'm done.

It's now your turn.

I'd like you to complete the working in order to solve root 2 multiplied by x is equal to 3x plus 1.

Please give your answer in the form a plus b root 2.

Pause the video and do this now.

Welcome back.

How did you get on? Well, you should first of all have started by trying to gather the like terms. In other words, taking 3x away from both sides.

You can then factorise and divide, which is the final step seen on the screen.

Of course, that's not the correct form for your answer.

Luckily of course you know how to rationalise the denominator, so let's multiply by root 2 plus 3.

This will give us a numerator of 3 plus root 2 and a denominator of 2 subtract 9.

Well 2 take away 9 is just negative 7.

However, I can't leave this as one fraction.

I must write it in the required form, so I'll divide each term of the numerator by negative 7.

This leaves me with a result of negative 3/7 take away 1/7 multiplied by root 2.

Well done if you got to the final answer.

And remembered to put it in the correct form.

It's now time for your first task.

I'd like you please to solve each of the linear equations you can see on the screen.

Please give your solutions in the form a multiplied by root b.

Pause the video and do this now.

Welcome back.

Let's see how you got on.

For a, we needed to remember, of course, to gather like terms, but what I've done first is I've simplified.

I know that in order to combine surds, whether through addition or subtraction, I need the radicand to be the same.

One of the useful ways to see if I can get the radicand to be the same of course is to simplify, and that's what I've done in my second line of working.

Root 45 has become 3 root 5, and the square root of 20 has become 2 root 5.

I've then gathered my like terms by subtracting 2x root 5 from both sides and adding 15.

This has got me to x multiply by root 5 equals 15.

I can then of course divide by root 5 but that's not of the desired form.

I need to rationalise that denominator.

By multiplying the fraction by root 5 over root 5 I reach 15 root 5 over 5, but I can divide 15 by 5.

That's just 3.

So my final solution is x equals 3 root 5.

In b I've done a similar thing.

I've simplified the root 12.

Root 12 simplified of course is 2 root 3 but the coefficient already present was 3x so I multiplied to give me 6x instead.

So we have 6x root 3.

I then subtracted 2x root 3 from both sides and I took away 4 from both sides, leaving me with a result of negative 6 on the right hand side.

Now, 6x root 3 subtract 2x root 3 just leaves me with 4x root 3.

Well the nice bit is I divided by 4 first of all, and then I divided by the root 3.

You could of course have done this in one go and just divided by 4 root 3.

At this point I've tied it up as much as possible, but it's still not of the required form.

I'm going to multiply therefore by root 3 over root 3 and then tidy up.

Doing that produces a final result of negative root 3 over 2.

Now let's consider c.

In c, I'm going to start of course by expanding the brackets but I'm also going to simplify all the surds that I can see.

This leads me to the line 2x subtract 8 root 2 equals 10 root 2 subtract x.

Now the next step of course is to gather the like terms, so I'm gonna add x to both sides and add 8 root 2 to both sides.

This leads me to the line 3x equals 18 multiplied by root 2, and I'm just gonna divide by 3 to get my result, x equals 6 root 2.

It's now time for part two of our lesson, and in this part we're going to be solving quadratic equations.

Now, in the unit on algebraic manipulation you did solve quadratic equations.

We solved them by factorising, completing the square, and by using the formula.

We're going to be focusing on the formula and completing the square in this lesson so you can see how surds are a part of this.

Just to recap, we know that some equations may have surds as coefficients.

We saw that in the first part of our lesson.

This can happen of course with quadratics as well.

And of course some equations may have surd solutions if we want the answer to be exact.

Let's have a look at solving a quadratic equation by completing the square.

We're going to leave our answers exactly now rather than rounding.

We of course start with our equation.

To complete the square, remember, we're going to be halving the coefficient of x in order to write x subtract 3, that's inside our brackets, and when we square this, we know that the result would be x squared subtract 6x, which are the first two terms of our quadratic.

So good so far, but the final term would be plus 9, and that's not what I need for my original quadratic.

I want plus 4.

Well, that means I need to take away 5, because 9 take away 5 would gimme that final term of 4.

Now that I've completed the square, I'm going to solve for x.

So we add 5 to both sides, square root.

Now in this case, I do want to have the plus or minus sign here.

Remember, it's a quadratic equation, so I am expecting there to be two solutions.

I then add 3 to both sides.

And there's my answers in exact form.

Now your turn.

Quick check.

True or false.

The solutions to y squared minus 4y plus 2 equals zero are y equals root 2 plus or minus 2.

Is that true or false? Once you've made your selection, don't forget to justify your answer by selecting either a or b.

Pause the video and do this now.

Welcome back.

Which one did you go for? It is of course false.

The solutions are 2 plus or minus root 2, not the other way round.

It's now time for your second task.

What you have here are four quadratic equations, and that's your top row.

The other two rows are a mishmash, and what you need to do is to work out which completed square form goes with which quadratic, and then which set of solutions goes with each quadratic.

Pause the video while you sort this out.

Welcome back.

How did you get on? Here are our quadratics along with their completed square form and therefore their final solution.

This should be quite familiar to you if you've completed the unit on algebraic manipulation.

As you can see, in each case the number that appears inside the bracket along with our variable is half the coefficient of the a term in each case.

We then of course know that when we expand that if we square that term inside the bracket we then need to work out how to get back to the final term in each of our original quadratics.

It's then a simple case to rearrange for the solutions.

Now, I did say that we were not just going to complete the square.

We were also going to use the formula, and we'll do that now.

Here's our quadratic on the screen.

Let's substitute into our formula.

You can see the formula on the right hand side of the screen.

When we substitute in, we have 8 plus or minus the square root of 64 minus 16 all over 2.

We can then of course simplify what's under the radical sign.

64 take away 16 is 48.

Now I know I can simplify that surd.

I'll therefore have 8 plus or minus 4 root 3 all over 2.

I can divide the numerator 3 by 2, so that will give me 4 plus or minus 2 root 3.

And there are my two solutions.

I'd like you now to find the exact solutions to the equation 3a squared plus 10a plus 5 equals zero.

We're going to do this in stages though.

We'll start by having a look at what Jacob and Laura recommend we do.

Jacob is telling me that the only way to solve that quadratic is to use the formula.

Laura, on the other hand, is telling me that the only way to do it is to complete the square.

Who is correct? Why? Pause the video while you write down who is correct and why you think that.

Welcome back.

I bet you didn't fall for this.

You know that neither of them is correct because both methods will work.

Jacob therefore goes on to use the formula since he preferred that method and he's found the solutions to 3a squared plus 10a plus 5 equals zero.

However, well done, Jacob, he's tried to check his working and he's pretty certain there's a mistake there, but he hasn't spotted it.

I'd like you please to pause the video and identify where Jacob has made his mistakes.

Welcome back.

Did you spot where he's gone wrong? That's right.

Right up here when substituting into the formula.

He has not multiplied by 3 and he should have.

Remember, our formula there says that we are subtracting 4ac and the denominator should be 2a.

So in each case he's forgotten to multiply by 3.

I'd like you please to now fix Jacob's work and give me the correct solutions to the equation.

I've put Jacob's working here on the left so you can see what needs to be corrected.

Pause the video and do this now.

Welcome back.

Let's see what you've done to Jacob's work.

On the right here you can see that we've now multiplied by 3 in both numerator and denominator where we needed to.

We've then tidied it up the surd, because 100 subtract 60 is 40.

Root 40 can of course be written in equivalent form, 2 root 10.

And then we can see that we can tidy this up by dividing each term by 2, leaving me with negative 5 plus or minus square root of 10, all divided by 3.

Now, I could of course write this in a different way, but you'll notice that I wasn't told that I had to give the solutions in a particular form.

So personally, given that that's the simplest fraction, and my denominator is rational, I'm going to stop there.

It's now time for your final task.

What you can see here are three quadratic equations and then you can see the working where I've used the formula to reach the final solutions.

Unfortunately, in each case, there have been mistakes.

Please find and then correct those mistakes.

That might mean you need to recalculate the final answer, so don't be afraid to do so.

Pause the video while you do this.

Welcome back.

Did you find all the mistakes? Let's have a look at where they are.

In a, the formula starts with minus b, and I've just substituted in b, didn't I? I had 18.

There was no negative sign there at all.

Having put that back in, I've then updated my solutions because it originally said 9 plus or minus 5 root 3 and it should have said negative 9 plus or minus 5 root 3.

In b I have the sign under my square root sign incorrect.

Remember, it's b squared subtract 4ac.

In this case, I would've had b squared, so that's negative 14 squared, which is 196, and then I'm subtracting 2 times 1 times negative 14.

Ah, there needs to be some addition there.

So I've put the plus sign in.

I've then carried on down by simplifying to reach my final solution.

In c I've substituted into my formula correctly.

I've then simplified under the radical sign correctly.

What I hadn't done was simplify root 40 correctly.

It shouldn't say 4 root 10.

It should say 2 root 10.

Having done that, we then correct the final solution.

Let's summarise what we've done in our lesson today.

Answers to equations can be left in exact form when we are solving.

This could be particularly useful when we're solving quadratics by completing the square or by using the formula.

Well done.

You've worked really well today and I can see how hard you tried.

I look forward to seeing you in our next lesson.