2026: GCSE Maths Edexcel Paper 2 Walkthrough 2026

Hello everyone, this is Paul from Third Space Learning, and today we're going to go through our Mathematics Paper 2 calculator paper for higher tier. Now this one aligns with Edexcel, uh, for GCSE, and it's set number 5 in our exam paper series. You can download this and all the other sets of papers and indeed hundreds of other free resources at thirdspacelearning.com. Highly recommend it, there's some fantastic stuff on there. So this paper, calculator paper, it's 1 hour and 30 minutes in length. Now we're going to spend a bit longer than that today because we're going to go into it in a little more detail. We're going to break down the questions, we're going to talk about it, and then we're going to go through where those all-important marks come from. So what you'll need is a pen and a calculator and a ruler and pencil, all the usual lovely math stuff. Let's go through it. You ready? Okay, let's dive in. Number 1: work out the value of x. Now I can see here we've got a right-angled triangle, so I'm thinking it's probably either going to be Pythagoras or trigonometry. But because I've got this angle here and I need to work out the value of it, it's going to be trigonometry. So let's start by labeling the sides of our triangle. So the side opposite the angle is always called the opposite. The side that is the longest is called the hypotenuse, and the side that is next to the angle is called the adjacent. OK, next job. Let's write down our SOHCAHTOA triangles. So, SOH, CAH, TOA. Okay, now what I'm looking for is on the sides. What information do I have? Well, I can see I've got information about the opposite, and I've got information about the adjacent. So which one of those triangles has information on the opposite and the adjacent? Well, it's this one, isn't it? That's got O and A in it. So this is the triangle that I need. So let's write this down. We want to work out the value of x, so which one of these am I finding? Well, it's tan of the angle, so it's going to be t. Cover up t and I get o divided by a. Now, t is just tan, and it's always tan of the angle, so tan of x in this case is equal to the opposite, which is 6, divided by the adjacent, which is 11. Next job, we need to work out what x is, so we need to do the inverse of tan. So the inverse of tan is inverse tan, so it's tan to the minus 1 of 6 over 11. Now put that straight in the calculator and we get 28.6104596 7. Always a good idea to write down the whole calculator screen. Now it says to 3 significant figures, so let's do that. So that's the first significant, second, that's the third one. Let's draw a little line. So to 3 significant figures, this is going to be 28.6. Right, let's see where those marks come from. So we're going to get 1 mark for creating one of these trig equations. So it's going to be either one of these two. It's going to get you a mark. And then the final accuracy mark is for the final answer. Okay, number 2: find the reciprocal of 0.4. So what does reciprocal mean? It just means 1 over. So the reciprocal of 0.4 is just 1 over 0.4. Which is straight in the calculator, 2.5. One mark. Okay, Part B. It's x 5.3 to 2 significant figures. Complete the error interval for x. So what we need here is the upper bound and the lower bound. So a good way to think about this is we've got this number here, the 3, is in the tenths column. So to find the upper and lower bounds, that's half 0.1, that will get me 0.05. Add it on and take it away, so we're going to get 5.35 and 5.25. So those two are the upper and lower bounds of 5.3. So let's pop it on the answer line: 5.25 less than or equal to x is less than 5.35. Where do those marks come from? So this is a 2-mark question, so we're going to get 1 mark for the lower bound and 1 mark for the upper bound. Number 3: scatter diagram. Lovely. Okay, the scatter diagram shows the history and geography marks of 10 students in their recent tests. Describe the relationship between the students' marks in history and the students' marks in geography? Well, this is about correlation, so we can see that they are going up, and we can think about this by drawing a line of best fit if that's helpful. So let's draw one on. So we want to try and get a straight line going up. It's going to go through the middle of these points with about the same number on either side of it. So If I draw it about there, about that. Okay, that looks pretty good to me. So we can see here the points are going up, so to get 1 mark for Part A, we just need to write positive correlation, and that's 1 mark. Part B: another student scored 25 in the history test. Use the graph to estimate this student's score on the geography test. Okay, so what we need to do is think about— it's 25 in the history test So we're going to go to 25 on the line of best— on the history marks, which is here. And let's draw a straight line all the way up to our line of best fit. Now you use a ruler to make sure it's nice and straight and make sure it just touches the line of best fit. And then let's draw a line all the way across to here and let's see where that goes to. It's about there. Okay, so I get mine at about 30. Now, there will be a range of answers here depending on where that line of best fit is. So my answer is 30. So let's see where those marks come from. We're going to get a mark for an attempt to read off a number from the line of best fit, and then the answer range here is given between 26 and 36, so a nice big range. Now, we are right in the middle of that. So very happy, 2 marks. All right, Part C: Both tests were out of 40. Which test do you think was easier? Explain why. Okay, so let's look at this. Well, we can see that the geography marks tend to be a little bit higher than the history marks, so perhaps the geography test was easier. So we'll say geography Seems easier, as the marks are generally higher. So, 1 mark for identifying geography and 1 mark for stating the reason. 2 marks. Okay, number 4: bearings. Lovely question. The bearing of A from B is 121. So remember, with bearings, we need to have it going from the north line, we need to measure clockwise, and we also need to have them in 3 figures. So let's draw the bearing of A from B before I do anything else. So let's start with our north line. And remember, this is from A from B, so that means that it's going from B. So here's B and here's A. Let's put a north line in here. And because it's going from B to A, from B to A is 121. Okay, what is the bearing of B from A. So we have to go from A to B, and remember, we have to go clockwise because it's a bearing. So starting from the north line, we need to go clockwise all the way around to there. So this is the angle I need. How can we work that out? Well, maybe we start with this angle. I can work out this angle because these are parallel lines. So these are co-interior angles, and I know that co-interior angles add up to 180. So I can say 180 subtract 121 is 59. So this is 59. Remember, if you know something, write it on the diagram. Okay, now we can work out this angle here, and we know that angles in a circle add up to 360. So we can say 360 subtract 59 is 301. So that's the answer, 301. One mark for a correct sketch or correct labeling or working out this, and an accuracy mark for 301. Okay, number 5. Frankie invests £7,000 in a bank account. So Frankie gets 6% per annum compound interest per year. After n years, she gets this amount of money. Work out the value of n. Okay, so all we have to do here is just try a few different values. So it doesn't really matter where we start, so let's just pick one. Let's try and think of something that's going to be sensible. Let's say after 4 years, so let's say n equals 4. Now that's going to mean £7,000 multiplied by— now it's the multiplier we need here. Now it's 6% per annum, so it's going up every year. So it's 100% plus 6% is 1.06. So it's 1.06 to the power of n, to the power of 4. Straight in the calculator really, really carefully, and we get 8,837.34. So it's not bad, we're not, we're not far away, are we? So we need to go a bit higher, so we need another year. So let's say n equals 5, and let's do the same thing again. 7,000 multiplied by 1.06 to the power of 5. Straighten calculator, and we get 9,367.58. Look at that! That is exactly what I wanted. So that means that the value of n is just 5. Remember to show the examiner everything you're doing. Don't just put this straight in the calculator. Write it down first and then put it in the calculator, because then we can pick up those all-important method marks. So we're going to get 1 mark for trying 1, and we're going to get 1 mark for the correct answer. Okay, let's have a look at question number 6. So use your calculator to work out that. So let's start with the top line. We're going to need the cube root button on our calculator. Now we normally get that by pressing Shift and then square root. So Shift square root, and then maybe in a bracket we could write 1.4 squared plus tan 30. We want to make sure that all of that, all of this bit, is underneath the square root. And then if we press equals, we're going to get 1.363934374. Okay, now leave that on your calculator screen and then press divide by 0.91 for that one. And when we do that, we get 1.4 4988 28982. And there it is. You were going to get 1 mark for that and 1 mark for that. Next job: round your answer to 3 significant figures. So we want to round 1.4988 28982 to 3 significant figures. So that's the first significant figure, that's the second, and that's third. So let's draw a line to the right of it And that 8 is bigger than 5, so we're going to have to round up. So 1.49 rounded up is 1.50. And there's that one mark. Okay, question number 7. The points A, B, and C form a straight line ABC. The coordinates of A are (2, 3) and the coordinates of B are (6, 9). Right, let's visualize this. I'm going to draw a quick sketch just to help me, help me see what's going on here. So that's the y-axis, and then there's my x-axis. And let's put in A, B, C on this line. So let's draw a straight line. Let's start with A. So there's 2 and 3. Let's have that as A. And let's draw a straight line up here. So if that's A, B is at 6, 9. So let's put B— let's say, let's say about there, there. And B is at 6, 9. And then let's have C somewhere up here, because we know that the distance from A to B, that's 2 parts. And then from B to C, that's 5 parts. So let's think about this. A to B, B to C is 2 to 5. So let's think about what it would take to get from A to B. Well, if we look at our diagram, to get from A to B, we need to go 4 this way, and then we need to go 6 this way. So that would be the vector 4, 6. Now, from B to C in the ratio, this is 2 parts, this one would be 5 parts. So let's think about what that would be. I'm going to have to divide both of these by 2 and then multiply them by 5. So let's do that. 2 divided by— 4 divided by 2 is 2. 2 times, uh, 5 is 10, and 6 divided by 2 is 3. 3 times 5 is 15. So to get from B to C, we're going to need to go 10 to the right, and we're going to need to go 15 up like that. Great. Now the question says work out the coordinates of C, so I need to know what that is and what that is there. Right, well, the distance all the way to this point is going to be 10 plus 4 plus this distance here, which is 2. So 10 plus 4 plus 2, well, that's 16. Now let's do the other one. I want to know what this is, so it's going to be this distance, 15, plus 6, plus this distance here, which is 3. So 15 plus 6 plus 3 is going to give me 24. So the coordinates of C are just (16, 24). Lovely question, 4 marks, right? Let's see where they come from. So we're going to get our first mark from thinking about how we get from A to B. In 4 along and 6 up, so there's 1 mark. Second method mark then is thinking about how to get from B to C— 2 marks. Then third mark, well, we're going to see that by adding on the values to work out this, and then our final answer is 16 to 24. Right, question number 8. So we've got two similar triangles here, so that means that one is an enlargement of the other one, and we want to work out the perimeter of triangle B. Okay, well, we've got the three sides, but we don't know what x is, so we're going to have to work out x in order to work out the numerical value for the perimeter. So let's do that. We know they're similar, and we've got two corresponding sides here with 18 and 12. So we can work out the scale factor for the length. So the scale factor is just 18 divided by 12, which is 1.5. So that tells me that all the lengths on triangle B are 1.5 times bigger than the lengths on triangle A. So I can make an equation. I could say that for these two 1.5 multiplied by 2x plus 10, that is the same as 5x plus 9. So let's work this out. Let's multiply out this bracket like so, and we're going to get 3x plus 15 equals 5x plus 9. Then let's rearrange it and solve it. So if I take away 3x on both sides, we'll get 15 equals 2x 9. Take away 9 from both sides, we get 6 2x. Divide by 2, we get x 3. Okay, this is good news. So let's now work out what this one is. So that's going to be 5 lots of whatever x was, which was 3. So 5 lots of 3 plus 9. Well, 5 plus 3 plus 9 is going to be 15 plus 9, which is 24. So that's that one. And then let's do this one. So it's going to be 3— 3 lots of 3 plus 4. Well, 3 times 3 is 9, plus 4 is going to give me 13. And now I just need to add them together. So show the examiner what we're doing. I'm going to write down that perimeter is equal to 24 plus 13 plus 18, because this is where my method marks are coming from. Add them all together and we get 55. So it's 55, and it's centimeters because it is a perimeter. And that's 5 marks on that one. Not bad at all. So we're going to get 1 mark for the scale factor, 1 mark for multiplying it through here in an equation, and 1 mark for x, uh, 1 mark for substituting your value of x into these 2 values, and 1 mark for the final answer. Okay, right, number 9. The group frequency table gives information about the time taken for 80 swimmers to swim 100 meters. So there it is, you can see we've got 7 took between 60 and 90, 14 people took between 80 and 120, etc. Complete cumulative frequency table. Not a problem. So from 60 to 90, 90, that is just 7. Now from 60 to 120, well, that's going to be this one and this one all together, which is 21. From 60 to 150, it's all 3 of those added together, which is 44. Then the next one is at 27. Which is 71, and then add the next one and we get 80. Now that's good news because we're told there were 80 swimmers. So in total, 60 and 1— no, 210 was 80. Right, that's one mark and we have smashed it. Part B: on the grid opposite, down here, draw a cumulative frequency graph for this information. Okay, so remember, for cumulative frequency, we always plot the upper points. So we're going to start with— this goes from 60 to 90, and that was at 7. So here's 60 to 90, and I'm going to plot this at 7. Now the scale is got 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So at 90, I'm just going to plot at 7 like that. Okay, next one, 120 is 21. So 120 here, we're going to go all the way up to 21 like that. Next one, uh, at 150 it's going to be 44. So 150 all the way up to 44. Worth taking your time on this one. At 180, 71. So 180, first little cross at 71. And the last one, 210, is at 80. There it is. Right, now we need to draw a cumulative frequency curve. Now this needs to be smooth, so I'd recommend doing it with a pencil. No rulers here. We start at 60 because this starts at 60, and we're going to go to 210. So one smooth curve. So starting at 60, we're going to go up through there, through that one, through that one, through that one, through that one. There we go, not bad. 2 marks. So we're gonna get 1 mark for the points all, uh, correctly pointed— uh, plotted, and then 1 mark for a smooth curve. Right, next part says use a graph to estimate the interquartile range. Well, the interquartile range is just the upper quartile subtract the lower quartile. Now that's when we take our data and we split it up into quartiles, into quarters. So at 80, the lower quartile is going to be, uh, a quarter of 80. So a quarter of 80 is 20. So we're after the 20th value here, which is here. So let's take a straight line with our ruler draw it across to our line of best fit, and then draw a line straight down like that. So I get— I've got a recycler there— I get 120 for that. And the upper quartile, well, that's going to be 3/4 of 80, so that's 60. And let's draw a line across at 60 to get there, all the way down here like so. Look at that, absolutely lovely. I'm going to get here. Now I'm going to think about what the range is here, what the scale is here. So I've got between 150 and 180, that's 10. So that's a gap of 30. So each one of these is going to be 10— 30 divided by 10, it's going to be 3, isn't it? So 3, 6, 9, 12, 15. So 155 plus 15 is going to be 165. So the interquartile range is just 165 subtract 120, and we're going to get our value of— now we can put this straight in the calculator if you so want to— 45. Now you're going to get a small range of options in the mark scheme. So that is well within what we need it to be. The trick here is lovely smooth curve and then straight lines when you are, uh, working out your lower and upper quartiles. So that's going to be worth 2 marks. Where do they come from? So a lower quartile in the correct range given in the mark scheme, an upper quartile in the correct range, and we're going to get 45 as our answer overall. Tap. Okay, number 10. Number 10: probability. Amina has a pack of playing cards. However, some of the cards are missing, so Amina's going to pick a card at random. The probability that she picks a card from each suit is shown in the table. So Amina has 10 heart cards. How many cards does Amina have in total? Right, well, how should we— how should we tackle this? We've got 10 heart cards, so 10 out of a total number is going to be 0.5. So 10 cards represents a quarter of the pack. So what would the whole pack be? It's going to be 4 lots of that, isn't it? It's going to be 40. So there must be 40 cards altogether. 1 mark for something that suggests that 10 is a quarter of all the cards, and then 1 mark for the final answer. Right, now Amina has 13 diamond cards. Okay, complete the table. So that means that 13 cards out of 40 are diamonds, so 13 divided by 40 is 0.325. So let's write that in here as well. So that must be the probability. Now we need to complete the table, so we need to know the probability for spades now. But we know that if we add together all these probabilities, we get 1. So let's do that. Let's do 1 subtract 0.25, subtract 0.275, subtract 0.325, and what do we get in the calculator? We get 0.15. Pop that straight in the table and underline it so the examiner's super clear. Where do those marks come from? Well, we're going to get 1 mark for the correct diamond value and 1 mark for the correct spade value, and you'll notice I've marked them on the table because I I can give marks for that, so fill in absolutely everything that you can. Part C: Amina designs a game. To win this game, you must draw a club. So a full pack of cards contains 52 cards with 13 of each suit. Amina says that it is more likely that a player will win if they use her pack of cards than if they use a full pack of cards. Is Amina correct? Right, well, let's think about this. The probability in a full pack of cards of drawing a suit— drawing a club would be 13 out of 52, wouldn't it? So 13 out of 52 is a quarter, it's 0.25. So 0.25, well, that's, that's less than the probability of drawing a club from Amina's pack. Which is less than 0.275. So that means it is more likely that you will win if you're using Amina's pack. So we can say that. So it's more likely You will win using Amina's pack of cards. Right, 2 marks. Where do they come from? So we're going to get 1 mark for seeing this somewhere, so that's 0.25 or a quarter, and then 1 mark for a correct comparison statement and conclusion. So we're going to need to see that comparison. Really important. Okay, question 11. It says expand and simplify this lovely thing. This is a classic question, a triple bracket expansion, and then we need to simplify it. So the trick is pick any two you like to start with. So I mean, let's just pick the first two. Let's not make our lives too difficult. 2x 1, x 5. So let's expand this. 2x times x, 2x squared. 2x times negative 5 is negative 10x. 1 times x, 1x. 1 times negative 5 is negative 5. And let's just quickly bring those together. Negative x plus x is negative 9x. Subtract 5. Take your time here, don't mess up on the simplifying. Think about those number lines. So that is that bit. Next job, we need to multiply the blue bit by the final bracket. So let's write this out again: 2x² 9x 5 bracket 3x 2. So there's my blue part, and there is my Ripart. Let's expand. 2x² multiplied by 3x is going to be 6x³. 2x² multiplied by 2 is going to be 4x². -9x multiplied by 3x is going to be -27x². -9x times 2 is negative 18x. Just take your time with this, no mistakes happening here, not at all. Minus 5 times 3x, negative 15x. Uh, negative 5 times 2 is negative 10. Right, okay, we're nearly there. Let's now collect these like terms together. So we've got 6x x cubed— well, there aren't any more x cubes, that's by itself. Let's think about this then. How many x squares do I have? I've got that one, I've got that one. So 4x squared subtract negative 27x squared is going to be negative 23x squared. Okay, now let's think about these ones. So these are the x terms. So negative 18x subtract —15x. So think about the number line, that's going to give me an even more negative answer, -33x. And the -10 is just by itself. So there we go, there's our answer. So we're going to get 1 mark for correctly expanding any pair of brackets, so we did that over here. We're going to get a second mark for an attempt to expand the third bracket, so that was us here. And then the accuracy mark at the end is for that one. Just take your time with those questions, right? Number 12. Number 12 says the line of the equation L1 is that, the line of the equation L2 is that. Show that the lines are perpendicular. Okay, well, we know that the gradients of perpendicular lines multiply to give -1, so if we can show that here these lines must be perpendicular. So remember that— remember that y mx c for straight lines, where m is the gradient. So for L1, we've got y 3x 7, so the gradient here is 3. That was nice and easy. Let's look at L2. L2 is 3y x 1. So let's rearrange this to make y the subject. So let's subtract y from both sides. We get 3y -x 1, and then I'm going to divide everything by 3. So y -1/3x 1/3. So that means the gradient here is The coefficient of x, that's -1/3, which is that one there. So if these lines are perpendicular, what I should see is 3 multiplied by -1/3 will give me -1, and it does. 3 times -1/3 is -1. Therefore, they're perpendicular. So we could— we might say that. We might say The gradients of l1 and l2 multiply to give -1, so they are perpendicular. Lovely. 2 marks, Marshall. Right, where are we gonna get these marks from? So we need to find— one mark is for finding this gradient, and then the final mark is for showing a correct conclusion, which is all of that there. Which is nice. Question number 13, and it's an enlargement question. We're going to enlarge shape A by a scale factor of -2 from a center 0, 0. So that's the origin. So let's pop in our origin. This means that in my new shape, all the side lengths are going to be double the size. But because it's negative, we're going to do this in a special way. So let's start from the origin and then let's see how I get to, let's say, this corner. I'm going to go 1 to the left and 1 up. So because it's negative now, I'm going to times each of them by 2 and go in the opposite directions. So rather than going 1 to the left, I'm going to go 2 to the right, and rather than going 1 up, I'm going to go 2 down. So there is my new point. Okay, and then similarly with that one, I would go 1 to the left and 3 up, so I'll go 2 to the right and I'll go 6 down. Let's do this one, uh, rather than going 5 to the left and 1 up, I'm going to go 10 to the right and 2 down. Just going to be there. And this one, that one was 5 to the left and 5 up, so it's going to be 10 to the right and 10 down, that's gonna be there. This one, uh, that's 3 to the left and 5 up, so that's going to be 6 to the right and 10 down. And I might even be able to do my final one just by eye here. It's going to be that point there. And now all we have to do is join them up. So using a pencil and a ruler very, very carefully, go from there to there, there to there, there to there, there to there, there to there, there to there. Look at that! Okay, there it is. I've got my freshly enlarged shape by a scale factor of -2. So one mark for enlarging it by a positive 2, or any scale factor of 2, which we did. And then one mark for my fully correct answer. Right, number 14. So the price of furniture in London is £345. The price of furniture in Venice then is €360, and in New York it's $385. And we have some exchange rates, and we notice that they're both in terms of £1, so that's going to be interesting. In which city is the furniture the most expensive? Show how you decide. Right, so we're going to have to compare these amounts, aren't we? And because the exchange rates are given to us in £1 equals £1 equals, let's turn them all into pounds. So, London— well, London's already in pounds, that's very helpful. So, London is £345. Let's look at Venice. Venice is €360. So let's take the euros. That's going to be £1 is equal to €1.16. And I want to convert from euros to pounds. So my question is, what do I do to 1.16 to get 1? Well, I divide it by itself, don't I? I divide it by 1.16. So let's divide that by 1.16 and we get £310.34. Remember, I'm rounding to 2 decimal places because it's money. Let's look at New York. New York was in dollars, $385, and it was £1 is equal to $1.37. So to get from dollars to pounds, what do I do to 1.37 to get 1? Well, I divide it by itself, so I'm going to divide that by 1.37, and when I pop that straight into my calculator, I get £281.02. And now look at that, I can directly compare all 3. Which one's the most expensive? Well, it's London, isn't it? So I would say London is the most expensive. Make sure you write down this final statement. Show the examiner you know exactly what you're doing. Let's mark this now. So, 1 mark— we're going to get a correct— a method mark for a correct conversion from euros to pounds, and then 1 mark from dollars to pounds, and then a final conclusion. Without that final conclusion, we're going to lose an important mark. Question 50: A circle's radius has increased by 12%. Find the percentage increase in the circle's area. Ooh, lovely question. Right, let's think about this. So if I was increasing by 12%, that would be 100% add on 12%, so my multiplier would be 1.12. And the area of a circle is equal to pi multiplied by the radius squared. However, if we're going to increase the radius by 12%, that means that the radius r is going to be multiplied by 1.12. So that's going to be pi multiplied by 1.12 lots of r squared. Well, what is 1.12r squared times pi? Well, that's going to be 1.2544πr². So this is the multiplier. This is how much more the area has increased in terms of the percentage. So the percentage is this bit here. It's 25.44%. So we'll get 1 mark for noting that there's 1.12 and multiplying it in the area formula. We're going to get another mark for actually multiplying it out, and then final answer mark for the percentage. Question number 16: show the equation x³ 2x 1 0 has a solution between x 2 and x 3. Okay. Okay, lovely. To do this, what we have to do is substitute both of these values into our equation, and what we should see is one positive answer and one negative answer, because what that means is the graph has crossed the x-axis. So the y-value has gone from negative to positive, and where it crosses the x-axis, that means that y is 0. So that is a solution, or a root as we sometimes call it. So let's do this nice and carefully. x 2 and show the examiner that we're substituting in our values. So we've got 2 cubed minus 2 lots of 2 squared minus 1, and I get— well, that's negative 1, so that's a good start. And then let's do x 3. That's going to be 3 cubed minus 2 lots of 3 squared minus 1, and that gets me 8. So that's good. So we've gone from a negative to a positive. So we can simply say change of sign shows root. 2 marks. So 1 mark is substituting 2 and 3— we have to do both of them here— into the equations. That's one mark. And then a correct conclusion— so we have to write down the statement— gives me my second mark. Part B: show that that equation, which is the same one up here, can be written in that format. So we're making one of the x's the subject. The question is, which one? Well, let's look at this. We can see we've got x equals the cube root of something. So in order to get that, that means we're going to have to make this one the subject, because then I can cube root it at the end. So x³ 2x² 1 0. So let's make x³ the subject. So let's add 2x² and 1 to both sides, and we're going to get left on the Left-hand side is x³, and on the right-hand side is 2x² 1. And then I'll just cube root both sides to give me x √(2x² 1). And there's my mark. Part C: starting with x₀ 2.5, use the iterative formula— there— 4 times to find an estimate for the solution of our equation and give each value to 4 decimal places. Okay, these are great questions, but we need to be careful here. So we're told that x0 is 2.5. Let's do x1. So x1, using this formula, is going to be the cube root of the previous x value times 2. So it's 2 lots of 2.5 squared plus 1. Pop it in the calculator and we're going to get 2.3811. Now we need to do x2, so we can do this again. We can do the cube root of 2 lots of the previous x value, which is 2.3811 squared plus 1, and we're going to get 2.310. 8. Lovely. Now I need to do this 4 times, so I don't know about you, but I'm getting a bit tired here. There's a lot of fiddly things going on. So this will get you the answer. You can keep on doing this. x3, x4 is what I'm after. But let me show you an even better way. So at the start here, I want you to type in your calculator 2.5 and press equals, okay? Then For x1, I want you to do this. I want you to press the cube root button, so Shift square root, and then inside those brackets you're going to write 2 bracket, and then you're going to press answer, so ANS, squared, close bracket, plus 1, and that equals 2.381. Exactly the same as before, no change. But look what happens now. If we press the equals button again, we get straight to this. If we press it again, it's going to use the answer and plug it back into the formula and to give me 2.2689. And if I press it a fourth time, I'm going to get 2.2437. How great is that? So it's going to save me a lot of faff and a lot of potential for messing up here. It's really important that you show at least one substitution because you're going to get a mark for that. So let's see where those marks come from. We're going to get one mark for correctly substituting 2.5 into our formula. So that's this one here. The rest of them are accuracy marks. So I get an accuracy mark for this one. And I get an accuracy mark for this one. 3 marks. Right, question number 17. So we've got a triangle and we have to work out the length of XY. So that is going to be this length. Okay, so when I'm thinking about triangles, I'm looking at— is it going to be some trigonometry perhaps? To, to find this missing side. Can't use SOHCAHTOA because SOHCAHTOA is only for right-angled triangles. So this is going to be the cosine rule because we've got an angle and then we've got two sides that enclose it. I want to find that side opposite that angle. So let's start by writing down the cosine rule. Well, the cosine rule says that a squared equals b squared plus c squared minus 2bc cos A. Okay, let's label our triangle. So this is the angle, which is 72, so I'm going to relabel that as capital A, which means the side opposite is little a. And then the other two, it doesn't really matter, so let's go with this one being capital B, which means the opposite side is little b. And let's go with this one being capital C, which means the opposite side is little c. Next, let's substitute those values into our cosine formula. So a squared— well, I want to find a, so a squared equals b squared, which is 8.1 squared, plus c squared, 3.11 squared. Minus 2 lots of b, times my c, multiplied by cos of the angle A, which is 72. There it is. And then let's pop that straight in the calculator really carefully, and we get 136.731349. And let's take the square root, so we get 11.69 3, 2, 1, 8, 0, 8. Always a good idea to write down the whole calculator screen. And then it says round to 3 significant figures, so let's do that. 1, 2, 3, so it's going to be 11.7. Let's pop that on the answer line, and let's just double-check that the answer is sensible. 11.7 for the side XY. Yeah, that's about right, isn't it? That works. Now it says work out the area of the triangle, right? So area of a triangle would normally be half times base times height, but we don't have the vertical height of the triangle here, so we can use a different formula. And that formula is half AB sine capital C. So let's just do a quick sketch of the triangle here to help us visualize it. So that's one side, and that's the other side, and that's the other side. And we know that that angle is 72 degrees. So we need to work out the area. So I need an angle C. So let's call that angle C, which means that that one is little c. And then for the other two, it doesn't matter. So let's stick to what we had before. That one was B, and then this time that one is going to be a. So here b, we've got that's 8.1, and c— sorry, and a is 11.3. So let's just substitute those values in. So we get 1/2 multiplied by a, 11.3, multiplied by b, which is 8.1, multiplied by sine of the angle, which is 72 degrees. Straight into the calculator, and the area is 43.52510147. Always write down the whole thing. Get it to the screen. And now it must be just 3 significant figures, so 1, 2, 3, draw a line. It's going to be 43.5 centimeters squared. Let's see where these marks come from. So for the top one, we need— got 1 mark for a correct substitution into the cosine rule. We got a second mark for working that out, and then a final accuracy mark. And for part B, we got again 1 mark for the correct substitution into the formula, and then a final mark for the answer. Okay, 18. Let's express AC in terms of a and b. So we've got some vectors here, this is good. So we're going to go from A to C, so from there to there, right, that way. So I need to work out a way of getting from here all the way to here in terms of little a and little b. So there's a few different ways we could do this. We could go this way, we could go from A to D because we know what that is, that's A 3B, and then we could go all the way up here, D to C. So that might be quite useful. So let's write that down. A to C is equal to AD plus DC. Okay, well, we know what AD is, so that's good. DC, on the other hand, we need to think about a little bit more. Now, we're told this— we're told that CE, CE, which is this bit, is in the ratio 5 to 4 with ED. So if we worked out the vector for this part here, we could use the ratio to work out the vector of the whole thing. So let's have a go at that. Let's have a go at going from D to E. So D to E, well, Well, to do that, we need to go— starting at D, we're going to go from D to A, and then we're going to go from A to E. So D to A plus A to E. Now, what is D to A? Well, that's the opposite of A plus 3B, so that's going to be negative A —3b, and then A to E is +3a 4b, like that. Remember, we had to put this one as negative because this is going from D to A, not A to D. So we've got this part. This part is negative a minus 3b plus 3a plus 4b. Let's bring that together. So let's simplify that. So I've got negative a add 3b— negative a add 3a is going to be 2a, and then negative 3b add 4b is going to be plus positive b. So that's helpful. So we know that this part from here to here is 2a b. Now, we want to take the whole thing now all the way up from D to C. Well, to get from D to C, what we're going to have to do is multiply it— multiply this section by 9 parts out of 4. Now, the reason we're going to do that is because This part, D to E, represents 4 parts, and the whole thing represents 9 parts. So we're going to do 9 over 4 multiplied by 2a plus b. Let's expand those brackets. So 9 over 4 multiplied by 2a, that's going to be 9 over 2a, and then plus 9/4 multiplied by b is going to be 9/4b. Okay, so this is looking pretty good. So we now know all the way from D to C. So let's put everything together. To go from A to C— I'll write it here— We wanted to do AD, which was this one, so it's going to be a 3b, and then we wanted to add to it DC, which is this one. Okay, so let's just bring this together. So we're going to have a 9/2a altogether is 11, and over 2a, and 3b plus 9 over 4b, altogether is 21 over 4b. And there we go, we have correctly worked out the vector a to c. Now this is with 4 marks, and it's quite, quite a lot going on here, so let's see where those marks come from. We're going to get our first mark for a correct expression for D to E. Then we're going to get another mark for using that ratio to work out that whole length D to C— whole vector D to C. We're going to get another mark for adding them up along a correct pathway, so in our case it was AC equals AD plus BC. And then our final mark is an accuracy mark, and that's for our correct answer. Okay, let's have a look at number 19. Right, 19 says write 11 12x 2x² in this form, where a, b, and c are integers. So we spot straight away this is a completing the square question. So let's do it. Lovely question, this. This is our quadratic expression. And what we're going to do is I'm going to rewrite this in the form negative x squared— negative 2x squared plus 12x plus 11. It's exactly the same thing, I'm just putting them in order of powers. So squared, power of 1, no power at all. Next job, we're going to take out a factor of 2 for this part here. So what we're left with is x² 6x. So that bit is that bit in there, and the add 11 is still on the outside. Then we're going to write the completed square form of x² 6x. So to do that, we're going to have a little bracket. We're going to do x subtract— well, half of -6 is -3. 3, so negative 3. We're going to square that. 3 squared is 9, take away the 9, and that plus 11 is still there. So that part is still exactly the same. Okay, let's now expand those brackets. So I'm going to multiply this 2 by each of these terms. So negative 2 multiplied by x minus 3 all squared is just negative 2 bracket x minus 3 all squared. Then we've got negative 2 times negative 9 is positive 18, and then the plus 11 is still there. And finally, bring it all together. 18 plus 11 is 29. And there it is. Now that will get you full marks, but Just so we can see where a, b, and c come from, let's just rewrite this so that the a is at the front. So it's going to be 29 subtract 2 lots of x take away 3 squared. And there we go, completely done. Just take your time with those questions, one thing at a time. Where do those marks come from? So we're going to get 1 mark for factorizing out negative 2. 1 mark for completing the square, and then 1 mark for the correct answer. Right, Part B. Now, Part B tells us that curve C has this equation, the one that we've just been rearranging. Point A is a maximum point of C. So let's just see what this would look like. This is a negative quadratic, so the quadratic— let's just say, let's just take some axes. It would look something a little bit like that, just very roughly. So point A is the maximum point. Use your answer to the first part to write down the coordinates of point A. Okay, so what we need to think about is the maximum point here. What that look like? So let's take our completed square form. We've got y 29 2(x 3)². So what we want here is for y to be as big as it possibly can be, the biggest point of y. So in order for that to be true, I've got 29 take away something. This part here is a squared term. Now, the smallest possible thing that, that this can be is 0, because when we square a number, we always get a positive answer. So I want to take the smallest possible value I can from 29 to keep it as big as I possibly can. So the question is then, what would this bracket need to be in order to be as small as it possibly could be? Well, we know that the square term can never be negative, so that would be 0 at its smallest possible term. Now if that is 0, that would be 29 minus 2 times 0, which is 29. So the y-value at the maximum point will be 29. So the question now is, well, what would the x-value be? So if this bracket, x 3, all squared is 0, what value of x will give me that? Well, I can just work it out. I can square it both sides, so we get x 3 0. Square root of 0 is 0. Add 3 to both sides, x 3. So x is 3 there. So at the maximum point we've got 329, and that's worth 1 mark. Lovely. Question number 20: show that that expression can be written in that form. Okay, we can do that. Where n, m, and p are expressions in terms— expressions in terms of a and b, and and states n, m, and p. Okay, a few things to do here. Before I do anything, I can simplify this a bit. I've got 1 plus the square root of 9a over b minus the square root of 4a. Well, the square root of 9 is just 3, so I can write that as plus 3, and the square root of a is just a, so plus 3 root a over b minus— well, the square root of 4 is 2, so it's gonna be 2, and the square root of a is root a. Okay, so that's looking a bit nicer. Now I need to get it looking like this, so I'm going to have to get rid of this surd on the bottom. So we need to rationalize the denominator. So how do we do that? Well, Let's start by writing this out again. So 1 3√a over b 2√a. And what we're going to do is multiply the top and the bottom by the denominator, but we're going to change that negative sign to a plus. We're going to change whatever that sign was. So here's negative, now it's a plus. I will do the same on the top. The reason we're doing that is because what we're effectively doing there is multiplying it by 1. We're multiplying the same— anything divided by itself is 1, so we're not changing the fraction at all. We're just going to change how it looks. Okay, so let's write this out using double brackets. 1 3√a in a bracket, and b 2√a a, and over b minus 2 root a and b plus 2 root a. Okay, let's start to expand out these brackets. So 1 times b is b. 1 multiplied by, um, 2 root a is plus 2 root a. 3 times— root a times b is going to be 3b root a. Be very careful of the order of that. 3 root a times 2 root a— well, 3 times 2 is 6, and root a times root a is just a. Divide by b times b is b squared. b times 2 root a is plus 2b root a. Then negative root a times b is going to be negative 2b times by root a. And negative 2 root a times 2 root a is going to be, well, negative 2 times positive 2 is negative 4. And root a times √a is going to be positive a. Okay, top line— can I simplify any of this down a bit? So I've got b. Okay, what do I— what have I got here? I've got 2√a and I've got 3b√a. So I've got a common factor of √a there, haven't I? So let's bring that out. Let's open a bracket. I can put √a there, and in this bracket is going to be 2 from here and +3b from here. If I were to expand this bracket out, I would get back to that. And then the 6a, I can't do anything with that, so pop that on at the end. Over— okay, what do we have? Well, we've got b², that's good. Here— oh, this is good— that take away that cancels. That becomes 0, and I'm just left with -4a. Right, this is looking pretty good. So let's see what the order I wanted it in. I wanted m n√a/p. So that's what I've got. Let's just rewrite this so it's a little bit clearer, and I'll do it up here. I'm gonna put my m value first. That's going to be b 6a 2 3b times by √a. So that's just this line, and the bottom line is staying exactly the same as b² 4a. So let's think about this. m is that part, so m b 6a. n is that part, so it's going to be 2 3b. And then last but by no means least, p— is that part? So p is b² 4a. And there we go. We have rewritten this in the correct form, and expressions n, m, and m are all in terms of a and b. So that's all done. That is worth 5 marks So let's see where those marks come from. So we're going to get 1 mark for attempting to multiply by this. We're going to get another mark for simplifying it down. I'm going to get an accuracy mark for reaching this line here, so that's good. We're going to get a mark for grouping the terms together, and then a final mark for clearly stating M n, and p. And there it is, 5 marks. Right, question number 21. So we've got a circle with this equation: x² y² 25. There it is. And we've got a line with the equation y 2x 5, and they're going to intersect at the points A and B. We want to know the length of that line. Right, what a good question this is. Let's start by doing a quick sketch just to help us visualize it. So let's draw a lovely pair of axes like that. I'll use a ruler for this. And let's draw a circle. So what's the radius of our circle? Well, x² y² r². That's the general format of a circle. So x² y² 25. r² is 25, isn't it? So r must be 5. So let's draw a circle with radius 5. So here we go, see how well I do this. There we go. And so that's going to be 5, 5, -5, -5. Okay, next let's, let's plot this one. I'll do it in blue. So this is a straight line, so this is y mx C, where C is the y-intercept. So here C is -5, so we know it's going to cross here, and we know it's a straight line, and we know it's a positive gradient— the gradient's 2. So let's draw something a bit like that, again with a ruler, just, just to give us a bit of an idea. Okay, and then let's put this point as A, and and this point is B, and I want to work out the length of that line. So I'm going to need to know what those coordinates are. So how to do that? Well, I can solve these equations simultaneously. I'm going to substitute this one into here. So we're going to have x squared plus y squared, which is 2x minus 5, all squared, equals 25. So let's start now to expand the brackets. So if I expand out x— 2x 5, double brackets there— expand out very carefully, I get 4x² 20x 25 25. Let's bring these together. I've got x² 4x², that's going to be 5x². -20 is happily sitting there by itself, and then let's subtract 25 from both sides, so I get 0. Great. Now to solve this, let's factorize it. So what do I have in common on both? I've got a common factor of 5x in both of those terms. So 5x outside, inside we're going to have x, subtract 4, equals 0. So this means that either 5x 0, or x 4 0. So if 5x 0, then x is just 0, or if x 4 0, then x 4. Okay, this is good news. So that's one, and that's another one. Now we need to work out what the y-coordinates are. Well, if we look at our diagram here, we can see that for A, the x-coordinate is 0. So A must have the coordinates (0, -5). So that's— that one's done for us already. y -5 there. I'll do that one in blue. But for this one, that's the x-coordinate of B, so we need to figure out what that is. So that's 4. And I need to work out the y-coordinate now. So let's substitute it in, and I can use either of my equations, doesn't matter in the slightest, but let's make our lives easy, let's use the linear one. Let's use this one. So y 2x 5, so that means that y 2 lots of 4 5, so that's 8 5, which is 3. So y equals 3. So that one's 3. Okay, so we have our coordinates now, so this is great news. We wanted to work out the length of AB, didn't we? So now we need to work out this bit. So let's think about this as a right angle triangle, and let's think about what this distance would be. Well, from x is 0 to x is 4, that's going to be 4. And let's think about what this distance would be. Well, that's from y is -5 to y is 3, so that's 8. So what I've effectively got there is a lovely little right-angled triangle like that. Well, that's 4 and that's 8, and that's A and that's B. So it's just Pythagoras. So we're going to do the length of AB is equal to 4 squared plus 8 squared square rooted, which is root 80, and that will simplify to 4 root 5. Remember, it says the exact length, so do not put a decimal there. If you're using a calculator there, which I would recommend, make sure it gives it to you in cert form. Now that is a 6-mark special. Let's see if we get all those marks. So we're going to get 1 mark for correctly substituting the y-value into the equation of the circle. We're going to get another mark for getting down to here. Then we're going to get a method mark for solving for x. Then we're going to get a method mark for solving for y. We're going to get another method mark for using Pythagoras, and then our final beautiful answer is 4√5.