Video: AQA GCSE Mathematics Higher Tier Pack 1 • Paper 3 • Question 26

AQA GCSE Mathematics Higher Tier Pack 1 • Paper 3 • Question 26

07:51

Video Transcript

Part a) Write four 𝑥 squared minus eight 𝑥 plus five in the form 𝑎 multiplied by 𝑥 minus 𝑏 all squared plus 𝑐. Part b) Use your answer to part a to solve four 𝑥 squared minus eight 𝑥 plus five equals 11.

In part a, we’ve been asked to write this quadratic expression in a particular form. And the form that we’ve been asked for is completed square form.

We need to determine the values of 𝑎, 𝑏, and 𝑐. There’s a procedure that we can follow for doing this. And first, we notice that the coefficient of 𝑥 squared is not one. It’s four. So our first step is going to be to factorise this four out, but only from the first two terms.

So we bring this factor four out. And inside the bracket, we then have 𝑥 squared minus two 𝑥, because four multiplied by 𝑥 squared gives four 𝑥 squared and four multiplied by negative two 𝑥 gives negative eight 𝑥. We then have the plus five as before.

Our next step is going to be to complete the square inside this bracket, so one 𝑥 squared minus two 𝑥. Again, this is standard process that we can follow. We always have 𝑥 plus or 𝑥 minus some number all squared. And to find this number, we can halve the coefficient of 𝑥.

So in the general form, if the coefficient was negative 𝑝, then in the bracket, we’d have negative 𝑝 over two. In our quadratic, the coefficient of 𝑥 is negative two. So inside the bracket, we have negative one. We then have to subtract this value squared. So in our case, we’re subtracting negative one squared. Negative one squared is positive one, so we’re subtracting one.

Let’s just confirm by expansion that 𝑥 minus one all squared minus one is indeed equal to 𝑥 squared minus two 𝑥. 𝑥 multiplied by 𝑥 is 𝑥 squared. 𝑥 multiplied by negative one is negative 𝑥. Negative one multiplied by 𝑥 is also negative 𝑥. Negative one multiplied by negative one is positive one. And then we still have the negative one on the outside of the brackets.

The positive one and negative one directly cancel each other out, which is what we wanted to happen. This is why we subtracted that value of negative one squared. The negative 𝑥 and then another lot of negative 𝑥 makes negative two 𝑥. And again, this is why we halve the coefficient of 𝑥, because we knew that when we expanded the brackets, we were gonna multiply this by 𝑥 twice, giving two lots of it. So by halving and then multiplying by two, we’re back to the original coefficient of 𝑥. We can see that our expansion does indeed give 𝑥 squared minus two 𝑥. So we’ve completed the square within the bracket correctly.

Now remember, all of this bracket was multiplied by four. So I’m going to put a big bracket around it and then the four on the outside. We then bring down the plus five as before. We’re nearly there, but the final step is just to expand that square bracket, the large bracket, so that we have this exactly in the form we were asked for. We have to multiply both 𝑥 minus one squared by four and negative one by four. And then we have the plus five as before.

It’s really important that you remember to multiply the constant term in that large bracket by four. A common mistake would just be to multiply four by the factor of 𝑥 minus one squared. This is why I drew in that large bracket so we’d remember that everything inside it needed to be multiplied by four.

Lastly, we just need to simplify. We have negative four plus five, which is equal to positive one. We’ve now written this quadratic in the requested form. The value of 𝑎 is four, the value of 𝑏 is one, and the value of 𝑐 is also one. In completed square form then, the quadratic four 𝑥 squared minus eight 𝑥 plus five is equal to four multiplied by 𝑥 minus one all squared plus one.

Now in part b, we’re told to use our answer to part a to solve this quadratic equation. And what you’ll notice is that the quadratic expression on the left of the equation is exactly the quadratic expression that we had in part a. So this suggests that we’re going to use our completed square form of this quadratic and substitute it into the given equation.

Replacing the left-hand side then with our completed square form gives four multiplied by 𝑥 minus one all squared plus one equals 11. And we now need to solve for 𝑥. Your first thought may be that we need to expand the bracket on the left-hand side. But actually we don’t need to. We can solve for 𝑥 by rearranging.

First, we subtract one from each side of this equation, giving four lots of 𝑥 minus one all squared is equal to 10. Next, we divide both sides of this equation by four. On the left, this will cancel with the factor four in the numerator, just leaving 𝑥 minus one all squared. And on the right, we have the fraction 10 over four, which can be simplified to five over two.

Next, we need to take the square root of each side of this equation. And in doing so, we must remember that whenever we solve an equation by square-rooting, there are two possible answers. So we need to take a plus or minus the square root. We have that 𝑥 minus one equals plus or minus the square root of five over two.

The final step is to add one to each side of this equation, giving 𝑥 equals one plus or minus the square root of five over two. And we have solved for 𝑥.

Now you may get the marks if you just leave your answer in this form. But what you’ll notice is that we actually have an irrational denominator, because we have the square root of a fraction. One of our rules of surds tells us that if we have the square root of a fraction 𝑎 over 𝑏, this is actually equal to the square root of the numerator over the square root of the denominator. So the square root of five over two is equal to the square root of five over the square root of two.

Root two is an irrational number. And whenever we have an irrational number in the denominator of the fraction, it’s good practice to rationalise this denominator. To do so, we multiply both the numerator and denominator of this fraction by root two. This is equivalent to multiplying by one, as both the numerator and denominator of what we’re multiplying by are the same. So we aren’t changing the size of the original fraction.

In the numerator, we have root five multiplied by root two, which is equal to root 10, because if we have the product of two square roots, this is actually equal to the square root of the product. So 10 is five times two. And in the denominator, root two multiplied by root two is equal to two. Two is a rational number. So we’ve now rationalised the denominator of this fraction.

We found that the solution to the equation four 𝑥 squared minus eight 𝑥 plus five equals 11 is 𝑥 equals one plus or minus root 10 over two. Now we could actually have got to this answer with a rational denominator a little bit more quickly if we hadn’t cancelled down at this stage here. If we’d left the fraction as 10 over four rather than five over two, then when we square-rooted, we’d have 𝑥 minus one equals plus or minus the square root of 10 over four.

Using our laws of surds, this would be equal to the square root of 10 over the square root of four. But four is a square number. Its square root is just two. So we could’ve replaced it here. We could then add one to each side, giving the same solution as before. 𝑥 equals one plus or minus root 10 over two. But we wouldn’t have needed to go through the process of rationalising the denominator.

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