Edexcel International GCSE Maths Solutions: 4WM1H/01 May 2025 (Unit 1 Higher)
Full worked solutions to the Pearson Edexcel International GCSE Mathematics A (Modular) Unit 1 Higher Tier exam paper from May 2025 (4WM1H/01, sat Thursday 15 May 2025): 25 questions, 100 marks, 2 hours, calculator allowed. Attempt each question yourself first, then click “Show solution” to check your working step by step.
Without using a calculator, work out 323 × 247
Show each stage of your working. Give your answer as a mixed number in its simplest form.
Show solution
- Step 1: convert to improper fractions. 323 = 113 and 247 = 187You can’t multiply mixed numbers directly. Multiply the whole number by the denominator and add the numerator: 3 × 3 + 2 = 11 and 2 × 7 + 4 = 18.
- Step 2: multiply. 113 × 187 = 19821 = 667 (or cancel the 3 into 18 first: 11 × 67 = 667)To multiply fractions, multiply the tops together and the bottoms together. Cancelling the common factor of 3 first keeps the numbers small.
- Step 3: convert back to a mixed number. 66 ÷ 7 = 9 remainder 3, so the answer is 937The question asks for a mixed number. 7 goes into 66 nine times (9 × 7 = 63) with 3 left over, so the 3 stays over 7.
The height of a tree is 17.8 metres correct to one decimal place.
(a) Write down the lower bound for the height of the tree.
The weight of an apple is 145 grams correct to the nearest 5 grams.
(b) Write down the upper bound for the weight of the apple.
A = 4.766 × 8150.399
(c) By rounding each number to one significant figure, work out an estimate for the value of A. Show your working clearly.
Show solution
- (a) One decimal place means ± 0.05: 17.8 − 0.05 = 17.75 mA value rounded to 1 d.p. can be up to half of 0.1 (0.05) away from 17.8. The lower bound is the smallest value that would still round to 17.8.
- (b) Nearest 5 grams means ± 2.5: 145 + 2.5 = 147.5 gRounding to the nearest 5 means the real value is within half of 5 (2.5) of 145. Anything from 142.5 up to 147.5 rounds to 145.
- (c) To 1 significant figure: 4.766 → 5, 815 → 800, 0.399 → 0.4
5 × 8000.4 = 40000.4 = 10 000Rounding each number to 1 s.f. turns it into an easy calculation. Dividing by 0.4 is the same as multiplying top and bottom by 10: 4000 ÷ 0.4 = 40 000 ÷ 4.
Here is a biased spinner. The spinner can land on pink or on green or on red or on yellow or on blue. The table gives information about the probability that, when the spinner is spun, it will land on each colour.
| Colour | pink | green | red | yellow | blue |
|---|---|---|---|---|---|
| Probability | 0.14 | 0.23 | 2x | 0.18 | 7x |
Jim is going to spin the spinner 500 times. Work out the expected number of times the spinner will land on blue.
Show solution
- Step 1: The probabilities add to 1: 0.14 + 0.23 + 0.18 + 2x + 7x = 1 → 0.55 + 9x = 1The spinner must land on exactly one colour, so the probabilities of all five colours add up to 1. Collect the numbers and the x terms separately.
- Step 2: 9x = 0.45 → x = 0.05Subtract 0.55 from both sides, then divide both sides by 9.
- Step 3: P(blue) = 7x = 7 × 0.05 = 0.35The table shows that blue has probability 7x, so substitute x = 0.05.
- Step 4: Expected number = 0.35 × 500 = 175Expected frequency = probability × number of trials.
The diagram shows triangle ABC and a semicircle.
AB = 15 cm, BC = 20 cm, AC = 25 cm, angle ABC = 90°. The diameter of the semicircle is 8 cm. Work out the area of the region shown shaded in the diagram. Give your answer correct to 3 significant figures.
Show solution
- Step 1: area of the triangle. The right angle is at B, so ½ × 15 × 20 = 150 cm²In a right-angled triangle, the two sides that meet at the right angle are the base and the perpendicular height, so area = ½ × base × height. The 25 cm side is not needed.
- Step 2: area of the semicircle. Radius = 8 ÷ 2 = 4 cm. ½ × π × 4² = 8π = 25.132… cm²A semicircle is half a circle, so its area is ½πr². The radius is half the diameter.
- Step 3: shaded area = triangle − semicircle. 150 − 25.132… = 124.867… = 125 cm² (3 s.f.)The shaded region is the triangle with the semicircle cut out, so subtract. Round to 3 s.f. only at the very end.
x5 × x7 = xm (a) Find the value of m
y8 ÷ y3 = yn (b) Find the value of n
(c) Simplify fully (5a4r2)3
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- (a) When multiplying, add the powers: 5 + 7, so m = 12x⁵ is five x’s multiplied together and x⁷ is seven more, so there are 12 x’s in total.
- (b) When dividing, subtract the powers: 8 − 3, so n = 5Dividing cancels the 3 y’s on the bottom with 3 of the 8 y’s on top, leaving 5.
- (c) Cube each part: 5³ = 125, (a4)³ = a12, (r²)³ = r6
Answer: 125a¹²r⁶The power outside the bracket applies to every factor inside. For a power of a power, multiply the indices: 4 × 3 = 12 and 2 × 3 = 6. Remember to cube the number 5 too.
The diagram shows a block in the shape of a cuboid.
The block has a horizontal base with length 6 m and width x m. The block is placed on a table so that the whole of its horizontal base is in contact with the table. The block exerts a force of 702 newtons on the table. The pressure on the table due to the block is 65 newtons/m². Work out the value of x.
pressure = forcearea
Show solution
- Step 1: rearrange the formula. area = forcepressure = 70265 = 10.8 m²If pressure = force ÷ area, then area = force ÷ pressure: swap the pressure and the area.
- Step 2: the base is a 6 × x rectangle. 6x = 10.8 → x = 1.8The area in contact with the table is the rectangular base: length × width = 6 × x. Divide both sides by 6 to find x.
(a) Solve x − 4 = 3 + 2x6 Show clear algebraic working.
(b) (i) Factorise y2 − 11y + 30
(ii) Hence solve y2 − 11y + 30 = 0
Show solution
- (a) Multiply both sides by 6: 6(x − 4) = 3 + 2x → 6x − 24 = 3 + 2x
Collect terms: 4x = 27 → x = 27/4 (= 6.75)Multiplying both sides by 6 clears the fraction. Put x − 4 in brackets so both terms are multiplied by 6. Then subtract 2x and add 24 to both sides, so the x terms are on one side and the numbers on the other. - (b)(i) Find two numbers that multiply to +30 and add to −11: −5 and −6.
Answer: (y − 5)(y − 6)To factorise y² + by + c, look for two numbers that multiply to c and add to b. The product is positive and the sum is negative, so both numbers must be negative. - (b)(ii) Set each bracket equal to 0: y = 5 or y = 6If two things multiply to give 0, one of them must be 0. So y − 5 = 0 or y − 6 = 0.
ℰ = {18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}
A = {even numbers} B = {multiples of 3} C = {multiples of 5}
(a) List the members of the set (A ∪ B)′
Sophie writes down the statement B ∩ C = Ø
(b) Explain why Sophie’s statement is wrong.
D is a set such that A ∩ D = {18, 26}. The set D has exactly 5 members.
(c) List the members of one possible set D.
Show solution
- (a) A = {18, 20, 22, 24, 26, 28, 30} and B = {18, 21, 24, 27, 30}.
A ∪ B = {18, 20, 21, 22, 24, 26, 27, 28, 30}
The complement is everything in ℰ that is not in A ∪ B: 19, 23, 25, 29∪ (union) means in A or B or both. The dash (′) means the complement: everything in ℰ that is not in the set. So the answer is the numbers that are neither even nor multiples of 3. - (b) C = {20, 25, 30}. The number 30 is a multiple of both 3 and 5, so B ∩ C = {30}, which is not empty.Ø is the empty set, and B ∩ C (intersection) means numbers in both B and C. To prove the statement wrong, give one number that is in both.
- (c) D must contain 18 and 26 and no other even numbers, so its other 3 members must be odd numbers from ℰ (19, 21, 23, 25, 27, 29).
One possible answer: D = {18, 19, 21, 23, 26}A ∩ D = {18, 26} means 18 and 26 are the only even numbers in D. The other three members must come from outside A, so they have to be odd.
7x = 1 (a) Write down the value of x
8−4 × 811812 = 8n (b) Work out the value of n
Show solution
- (a) Any non-zero number to the power 0 is 1, so x = 0By the index laws, 7ⁿ ÷ 7ⁿ = 7⁰. Anything divided by itself is 1, so 7⁰ = 1 and the power must be 0.
- (b) Numerator: 8−4 + 11 = 87. Then 87 ÷ 812 = 87 − 12 = 8−5
n = −5When multiplying powers of the same base, add the indices (−4 + 11 = 7). When dividing, subtract them (7 − 12 = −5). Take care with the negative numbers.
ABCD is a trapezium with one line of symmetry. Angle ADC = 60°, AD = 12 cm, DC = 47 cm. Work out the area of the trapezium. Give your answer correct to 3 significant figures. Show your working clearly.
Show solution
- Step 1: drop perpendiculars from A and B to DC, making a right-angled triangle at each end.The trapezium area formula needs the perpendicular height and both parallel sides. Neither is given, but the perpendiculars create right-angled triangles where trigonometry works.
- Step 2: height. h = 12 sin 60° = 10.392… cmIn the right-angled triangle at D, AD = 12 is the hypotenuse and h is opposite the 60° angle. SOH gives opposite = hypotenuse × sin.
- Step 3: horizontal overhang. 12 cos 60° = 6 cm at each end (the trapezium is symmetrical).The horizontal side is adjacent to the 60° angle, so CAH gives adjacent = hypotenuse × cos. Because the trapezium has one line of symmetry, the overhang at C is also 6 cm.
- Step 4: top side. AB = 47 − 6 − 6 = 35 cmThe top side AB is the bottom side DC with both overhangs taken away.
- Step 5: area. ½(35 + 47) × 10.392… = 41 × 10.392… = 426.08… = 426 cm² (3 s.f.)Area of a trapezium = ½(a + b)h, where a and b are the parallel sides. This formula is on the formulae sheet. Use the unrounded height.
(a) Expand and simplify 5x(5x + 4)(2x − 3)
(b) Express 94 + y − 75y as a single fraction in its simplest form.
Show solution
- (a) Step 1: (5x + 4)(2x − 3) = 10x² − 15x + 8x − 12 = 10x² − 7x − 12Expand two brackets first. Multiply every term in the first bracket by every term in the second, then collect like terms: −15x + 8x = −7x.
- Step 2: 5x(10x² − 7x − 12) = 50x³ − 35x² − 60xMultiply each term inside the bracket by 5x: 5x × 10x² = 50x³, 5x × (−7x) = −35x², and 5x × (−12) = −60x.
- (b) The lowest common denominator of 4 and 5y is 20y.
94 = 45y20y and y − 75y = 4(y − 7)20y = 4y − 2820y
Add: 45y + 4y − 2820y = 49y − 2820yTo add fractions they need the same denominator. 20y is the smallest expression that both 4 and 5y divide into. Whatever you multiply the bottom by, multiply the top by the same.
Ricardo is going to play one game of badminton and one game of tennis. He can either win or lose each game. The probability that he will win the game of badminton is 0.7. The probability that he will win the game of tennis is 0.4.
(a) Complete the probability tree diagram.
(b) Work out the probability that Ricardo loses both games.
Show solution
- (a) Badminton: win 0.7, lose 1 − 0.7 = 0.3. Tennis (on both pairs of branches): win 0.4, lose 1 − 0.4 = 0.6Win and lose are the only two outcomes, so their probabilities add to 1. The tennis result does not depend on the badminton result, so the same tennis probabilities go on both pairs of branches.
- (b) Multiply along the “lose, lose” path: 0.3 × 0.6 = 0.18On a tree diagram, ‘this AND then that’ means multiply the probabilities along the branches.
(a) Complete the table of values for y = x3 + 2x + 3
| x | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| y | −9 | 0 | 3 | 6 | 36 |
(b) On the grid, draw the graph of y = x3 + 2x + 3 for −3 ≤ x ≤ 3
Show solution
- (a) x = −3: (−3)³ + 2(−3) + 3 = −27 − 6 + 3 = −30
x = 2: 2³ + 2(2) + 3 = 8 + 4 + 3 = 15Substitute each missing x value into the equation. Put negatives in brackets: (−3)³ = −27, because a negative number cubed stays negative. - (b) Plot all seven points, (−3, −30), (−2, −9), (−1, 0), (0, 3), (1, 6), (2, 15) and (3, 36), and join them with a smooth curve.Each column of the table gives one point (x, y). Check the scale on the y-axis before plotting, because one small square is not one unit.
The diagram shows sector OAB of a circle, centre O. OA = OB = 12 cm, angle AOB = 40°.
(a) Work out the area of the sector. Give your answer correct to 3 significant figures.
The diagram shows triangle XYZ. XY = 7 cm, YZ = 9 cm, angle XYZ = 115°.
(b) Work out the area of the triangle. Give your answer correct to 3 significant figures.
Show solution
- (a) A sector is a fraction θ360 of the whole circle:
Area = 40360 × π × 12² = 19 × 144π = 16π = 50.265… = 50.3 cm²40° is 40/360 = 1/9 of a full turn, so the sector is 1/9 of the whole circle’s area, πr². - (b) Two sides and the angle between them are given, so use Area = ½ab sin C:
½ × 7 × 9 × sin 115° = 31.5 × 0.9063… = 28.548… = 28.5 cm²The triangle has no right angle, so ½ × base × height can’t be used directly. ½ab sin C works when you know two sides and the angle between them. Make sure your calculator is in degrees.
Solve the equation 5x2 + 9x − 17 = 0. Show your working clearly. Give your solutions correct to 3 significant figures.
Show solution
- Step 1: Use the quadratic formula with a = 5, b = 9, c = −17:
x = −9 ± √(9² − 4 × 5 × (−17))2 × 5 = −9 ± √42110‘Give your solutions to 3 significant figures’ tells you the quadratic won’t factorise, so use the formula from the formulae sheet. Write down a, b and c with their signs before substituting. - Step 2: √421 = 20.518…
x = −9 + 20.518…10 = 1.1518… or x = −9 − 20.518…10 = −2.9518…The ± gives two solutions, one with + and one with −. Work each one out separately and round only at the end. - x = 1.15 or x = −2.95Round each solution to 3 significant figures, as the question asks. A quadratic usually has two solutions, so give both.
Solve the equation 83y + 4 × 43y = 25y. Show your working clearly.
Show solution
- Step 1: write everything as a power of 2. 8 = 2³ and 4 = 2²
83y+4 = 23(3y+4) = 29y+12 and 43y = 22(3y) = 26yPowers can only be combined or compared when they have the same base, and 8 and 4 are both powers of 2. Then (aᵐ)ⁿ = aᵐⁿ, so multiply the whole of 3y + 4 by 3. - Step 2: add the powers on the left. 29y+12+6y = 215y+12When multiplying powers of the same base, add the indices.
- Step 3: equate the powers. 15y + 12 = 5y → 10y = −12 → y = −1.2 (= −6/5)If 2 to one power equals 2 to another, the two powers must be equal. That leaves an ordinary linear equation to solve.
(√3)5 = k√3 where k is an integer.
(a) Find the value of k
(b) Show that 213 − √2 can be written in the form c + √d where c and d are integers. Show each stage of your working clearly.
Show solution
- (a) (√3)5 = (√3)4 × √3 = 9√3, so k = 9(√3)² = 3, so (√3)⁴ = 3 × 3 = 9. Splitting the power 5 into 4 + 1 leaves one √3 on its own.
- (b) Step 1: rationalise by multiplying top and bottom by (3 + √2):
21(3 + √2)(3 − √2)(3 + √2)To remove a surd from the denominator, multiply the top and bottom by the conjugate: the same terms with the opposite sign. This doesn’t change the value, because you are multiplying by 1. - Step 2: denominator. (3 − √2)(3 + √2) = 9 − 2 = 7(a − b)(a + b) = a² − b² (the difference of two squares), and (√2)² = 2, so the surds cancel out.
- Step 3: simplify. 21(3 + √2)7 = 3(3 + √2) = 9 + 3√221 ÷ 7 = 3, then multiply out the bracket.
- Step 4: write 3√2 as a single surd. 3√2 = √(9 × 2) = √18
So 213 − √2 = 9 + √18 (c = 9, d = 18) ✓The form c + √d needs a single surd. Move the 3 inside the root by writing it as √9: √9 × √2 = √18.
The histogram shows information about the times, in minutes, that some people were in a shop.
Work out an estimate for the proportion of these people who were in the shop for more than 40 minutes.
Show solution
- Step 1: frequency = frequency density × class width for every bar:
Total number of people = 25 + 20 + 54 + 10 + 17 = 126In a histogram, the area of each bar (not its height) represents the frequency. Read the height from the frequency density scale and the width from the time axis.Time (min) 0–10 10–15 15–30 30–50 50–60 FD 2.5 4 3.6 0.5 1.7 Frequency 25 20 54 10 17 - Step 2: people with more than 40 minutes. Half of the 30–50 bar (40–50: 10 × 0.5 = 5) plus the whole 50–60 bar (17): 5 + 17 = 2240 minutes falls in the middle of the 30–50 bar. Assuming the people in that bar are spread evenly, the 40–50 part (half the width) holds half of its 10 people. That assumption is why the answer is an estimate.
- Step 3: proportion. 22126 = 11/63 (≈ 0.175)Proportion = part ÷ whole. Simplify by dividing the top and bottom by 2.
Simplify fully 3x2 − 3y25x + 5y ÷ xy2 − x2y10xy
Show clear algebraic working.
Show solution
- Step 1: factorise everything.Algebraic fractions can only be simplified after factorising, because only common factors cancel. Take out common factors first, then look for a difference of two squares: x² − y² = (x − y)(x + y).
- 3x² − 3y² = 3(x − y)(x + y)
- 5x + 5y = 5(x + y)
- xy² − x²y = xy(y − x) = −xy(x − y)
- Step 2: flip the second fraction and multiply. 3(x − y)(x + y)5(x + y) × 10xy−xy(x − y)Dividing by a fraction is the same as multiplying by its reciprocal, which is the fraction turned upside down.
- Step 3: cancel (x + y), (x − y) and xy: 3 × 105 × (−1) = −6Each of these factors appears on both the top and the bottom, so each one cancels to 1. What remains is (3 × 10) ÷ (5 × −1).
Given that k = m + n and m = 34n, express 7k5 − m in the form a + bn2cn − d where a, b and c are integers and d is prime.
Show solution
- Step 1: write k as a single fraction. k = 34n + n = 3 + 4n²4n, so 7k = 7(3 + 4n²)4nSubstitute m = 3/4n, then write n as 4n²/4n so both terms share the denominator 4n.
- Step 2: write 5 − m as a single fraction. 5 − 34n = 20n − 34nWrite 5 as 20n/4n so it can be combined with 3/4n over the same denominator.
- Step 3: divide. The 4n denominators cancel: 7(3 + 4n²)20n − 3When you divide two fractions with the same denominator, the denominators cancel, leaving top ÷ top.
- Step 4: expand the numerator. 21 + 28n²20n − 3 (a = 21, b = 28, c = 20, d = 3, and 3 is prime ✓)Multiply out 7(3 + 4n²) to match the required form, then read off a, b, c and d.
In the diagram, AB = 9.4 cm, AD = 12.8 cm, angle BAD = 72°, angle BDC = 39° and angle BCD = 54°. Work out the length of BC. Give your answer correct to 3 significant figures.
Show solution
- Step 1: find BD with the cosine rule in triangle ABD (two sides and the angle between them):
BD² = 9.4² + 12.8² − 2 × 9.4 × 12.8 × cos 72°
= 88.36 + 163.84 − 74.36… = 177.83…
BD = 13.335… cmBC is in triangle BCD, which needs side BD, and we don’t know it yet. Triangle ABD has two known sides and the angle between them, which is exactly when to use the cosine rule a² = b² + c² − 2bc cos A. - Step 2: find BC with the sine rule in triangle BCD. BC is opposite the 39° angle and BD is opposite the 54° angle:
BCsin 39° = 13.335…sin 54° → BC = 13.335… × sin 39°sin 54° = 10.373…In triangle BCD we now know side BD and its opposite angle (54°), plus the angle opposite BC (39°). A side with its opposite angle means the sine rule: a/sin A = b/sin B. - BC = 10.4 cm (3 s.f.)10.373… to 3 significant figures is 10.4, because the fourth figure (7) rounds the 3 up.
A box contains 20 counters. 9 of the counters are red, 7 of the counters are yellow, 4 of the counters are green. Alex takes at random three counters from the box. Work out the probability that exactly two of the three counters are the same colour.
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- Method: 1 − P(all three the same) − P(all three different). The counters are not replaced.There are only three possibilities: all the same colour, exactly two the same, or all different. Their probabilities add to 1, so it’s quicker to find the other two and subtract. Without replacement, the totals go 20, 19, 18.
- Step 1: P(all the same colour).For three reds: the first is red with probability 9/20, then 8 reds remain out of 19, then 7 out of 18. Do the same for each colour, then add the three results, because these are separate outcomes (‘or’ means add).
- RRR: 920 × 819 × 718 = 5046840
- YYY: 720 × 619 × 518 = 2106840
- GGG: 420 × 319 × 218 = 246840
- Step 2: P(all different colours). One order is 920 × 719 × 418 = 2526840. There are 3! = 6 orders, so the total is 15126840One red, one yellow and one green can come out in any order (RYG, RGY, YRG and so on). There are 3 × 2 × 1 = 6 orders, each with the same probability, so multiply by 6.
- Step 3: P(exactly two the same) = 1 − 7386840 − 15126840 = 45906840 = 51/76 (≈ 0.671)Subtract both probabilities from 1, then simplify by dividing the top and bottom by 90.
The diagram shows cuboid ABCDEFGH.
The point P lies on CF. AD = 12 cm, DE = 10 cm, CP = 3 cm. The angle of elevation of P from A is 24°. Calculate the size of angle APG. Give your answer correct to the nearest degree. Show your working clearly.
Show solution
- Step 1: the point below P. P is 3 cm along the 10 cm edge, so the point P′ directly below it on the base is 12 cm along and 3 cm across from A.
AP′ = √(12² + 3²) = √153 = 12.369… cmTo use the angle of elevation we need a right-angled triangle with a horizontal side and a vertical side. P′ is directly below P, and AP′ is the diagonal of a 12 by 3 rectangle on the base, so use Pythagoras. - Step 2: height of the cuboid from the angle of elevation: h = PP′ = √153 × tan 24° = 5.5071… cmThe angle of elevation is the angle between the horizontal AP′ and the line AP. In triangle APP′, PP′ is opposite the 24° angle and AP′ is adjacent, so use tan.
- Step 3: the three sides of triangle APG.Triangle APG is not right-angled, so finding its angle needs the cosine rule with all three sides. Each side is the hypotenuse of a right-angled triangle we can solve.
- AP = √153cos 24° = 13.539… cm
- PG lies in the top face: 12 cm along and 10 − 3 = 7 cm across, so PG = √(12² + 7²) = √193 = 13.892… cm
- AG is the diagonal of the end face ABGH (10 cm by h): AG = √(10² + 5.5071…²) = 11.416… cm
- Step 4: cosine rule for angle APG.
cos APG = AP² + PG² − AG²2 × AP × PG = 183.33… + 193 − 130.33…2 × 13.539… × 13.892… = 246376.20… = 0.6539…The rearranged cosine rule finds an angle from three sides. The angle at P is between AP and PG, so AG (the side opposite P) is the one subtracted. - APG = cos⁻¹(0.6539…) = 49.16…° = 49°Inverse cosine (cos⁻¹) turns the cosine value back into an angle. The question asks for the nearest degree.
PQRS is a square. PR is a diagonal of the square. P is the point with coordinates (4, 7). R is the point with coordinates (8, −5). Find an equation of the straight line that passes through the points Q and S. Give your answer in the form ay = bx + c where a, b and c are integers.
Show solution
- Key fact: the diagonals of a square bisect each other at right angles. So QS passes through the midpoint of PR and is perpendicular to PR.This means we never need to find Q or S. One point on the line (the midpoint) and its gradient are enough to write its equation.
- Step 1: midpoint of PR. (4 + 82, 7 + (−5)2) = (6, 1)The midpoint is the average of the x-coordinates and the average of the y-coordinates.
- Step 2: gradient of PR. −5 − 78 − 4 = −124 = −3Gradient = change in y ÷ change in x. Subtract the coordinates in the same order on the top and the bottom.
- Step 3: gradient of QS. Perpendicular gradients multiply to −1, so m = 13Perpendicular gradients multiply to −1, so the new gradient is the negative reciprocal: flip −3 to −1/3, then change the sign to get 1/3.
- Step 4: equation through (6, 1). y − 1 = 13(x − 6) → 3y − 3 = x − 6Use y − y₁ = m(x − x₁), then multiply by 3 to clear the fraction and rearrange into the form ay = bx + c.
- 3y = x − 3This is the required form ay = bx + c, with a = 3, b = 1 and c = −3.
(a) Express 8 − 12x − 4x2 in the form a − b(x + c)2 where a, b and c are constants.
(b) Hence solve 8 − 12x − 4x2 = 0. Show your working clearly. Give your solutions in surd form.
Show solution
- (a) Step 1: factor −4 out of the x terms. 8 − 12x − 4x² = 8 − 4(x² + 3x)Completing the square is easiest when the x² coefficient is 1, so take −4 out of the x terms. Check: −4 × x² = −4x² and −4 × 3x = −12x.
- Step 2: complete the square. x² + 3x = (x + 32)² − 94Halve the coefficient of x (3 ÷ 2 = 3/2) to make the bracket. Squaring the bracket adds an extra (3/2)² = 9/4, so subtract it.
- Step 3: substitute back. 8 − 4[(x + 32)² − 94] = 8 + 9 − 4(x + 32)²
17 − 4(x + 1.5)² (a = 17, b = 4, c = 32)Multiply −4 into the square bracket: −4 × (−9/4) = +9, and 8 + 9 = 17. - (b) 17 − 4(x + 32)² = 0 → (x + 32)² = 174 → x + 32 = ±√172
x = −3 + √172 or −3 − √172‘Hence’ means use part (a). Rearrange so the squared bracket is on its own, then square-root both sides. Use ± because both a positive and a negative number square to 17/4, and √(17/4) = √17/2.
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