Flashcards · 9260 · Topic 2

Algebra flashcards

56 cards on Topic 2, subtopics 3.2.1 to 3.2.4. Pick a subtopic or study them all.

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3.2.1 Notation and manipulation

Expression, equation and identity: Expression: no equals sign (3x + 2). Equation: true only for specific values. Identity: true for all values.
Expand (x − 2)(x + 3): x² + 3x − 2x − 6 = x² + x − 6
Factorise x² + 5x + 6, and x² − 9: Two numbers with product 6 and sum 5: (x + 2)(x + 3). Difference of two squares: (x − 3)(x + 3).
Simplify x/3 + x/4: Common denominator 12: 4x/12 + 3x/12 = 7x/12.
Make r the subject of A = πr²: Divide by π: r² = A/π. Square root: r = √(A/π).
Why can x²y³ and x³y² not be collected as like terms?: Like terms need identical powers of every letter; here the powers of x and y differ. (Multiplying them is fine: x²y³ × x³y² = x⁵y⁵.)
Factorise 2x² + 5x + 3Extension: ac = 6; two numbers with product 6 and sum 5: 2 and 3. 2x² + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1)
Simplify (2x² + 3x)/(4x² − 9)Extension: Factorise: x(2x + 3) / ((2x − 3)(2x + 3)). Cancel the common factor (2x + 3): x/(2x − 3)
Changing the subject when it appears twice: make x the subject of ax + b = cx + dExtension: Collect the x terms: ax − cx = d − b. Factorise: x(a − c) = d − b. Divide: x = (d − b)/(a − c).
Expand (x + 1)(x + 2)(x + 3)Extension: First two brackets: x² + 3x + 2. Then × (x + 3): x³ + 6x² + 11x + 6.
Prove that the sum of two consecutive odd numbers is a multiple of 4Extension: Write them as 2n + 1 and 2n + 3, where n is an integer. Sum = 4n + 4 = 4(n + 1). n + 1 is an integer, so the sum is a multiple of 4.

3.2.2 Functions, graphs and calculus

Function machine: write "add 3 then multiply by 5" and "multiply by 5 then add 3": Add 3 then multiply by 5: 5(x + 3). Multiply by 5 then add 3: 5x + 3. To find an input from an output, undo the steps in reverse order.
Domain and range of a functionExtension: Domain: the allowed inputs (1/(x − 4) needs x ≠ 4). Range: the resulting outputs.
f(x) = 2x + 3 and g(x) = x² − 5. Find fg(x).Extension: fg(x): do g first, then f. fg(x) = 2(x² − 5) + 3 = 2x² − 7
h(x) = (3x − 4)/5. Find h⁻¹(x).Extension: y = (3x − 4)/5, so 5y = 3x − 4 and x = (5y + 4)/3. h⁻¹(x) = (5x + 4)/3. Check: h(h⁻¹(x)) = x.
Find the gradient and y-intercept of 3x + 4y = 12: Rearrange to y = mx + c: 4y = −3x + 12, so y = −(3/4)x + 3. Gradient −3/4, y-intercept 3.
Gradient of a line perpendicular to a line with gradient mExtension: −1/m, because m₁ × m₂ = −1. Example: perpendicular to gradient −7/3 is gradient 3/7.
Find the equation of the line through (−3, −4) and (5, 8)Extension: Gradient = (8 − (−4)) ÷ (5 − (−3)) = 12/8 = 3/2. Substitute (5, 8): 8 = 7.5 + c, so c = 1/2. y = (3/2)x + 1/2
Find the roots and turning point of y = x² + 6x + 8: (x + 2)(x + 4) = 0: roots −2 and −4. Axis of symmetry halfway: x = −3. y = 9 − 18 + 8 = −1, so the minimum is (−3, −1). y-intercept 8.
Turning point from completed square form: x² − 10x + 21Extension: (x − 5)² − 25 + 21 = (x − 5)² − 4. y = (x − p)² + q has a minimum at (p, q), so (5, −4).
Shapes of y = x³ and y = 1/x: y = x³: S-shaped, rising, through the origin, no turning point. y = 1/x: two branches in quadrants 1 and 3, never touching the axes (x ≠ 0).
Graph of y = kˣ (k > 0)Extension: Always passes through (0, 1) and stays above the x-axis. Rising if k > 1, falling if 0 < k < 1.
Solve sin x = 0.8 for 0° ≤ x ≤ 180°Extension: sin⁻¹ 0.8 = 53.1°. The second angle is 180° − 53.1° = 126.9°. sin and cos have period 360°; tan has period 180°.
A water bill line passes through (0, 18) and (20, 66), cost in £ against cubic metres. Interpret it.: Gradient = 48/20 = 2.4: £2.40 per cubic metre. Intercept 18: a fixed charge of £18.
Velocity–time graph: what does the gradient show?: Gradient: acceleration (a downward slope means slowing down; a horizontal line means constant velocity).
A lift goes from rest to 3 m/s in 4 s, holds 3 m/s for 10 s, then stops in 2 s. Find the distance.Extension: Area of the trapezium: parallel sides 10 s and 16 s, height 3 m/s. ½ × (10 + 16) × 3 = 39 m
y is proportional to x³, and y = 40 when x = 2. Find y when x = 3.Extension: y = kx³: 40 = 8k, so k = 5. y = 5 × 27 = 135
Differentiate y = 4x³ − x² + 6Extension: Multiply by the power, then reduce the power by 1; a constant gives 0. dy/dx = 12x² − 2x
Find the equation of the tangent to y = x² − 4x + 7 at x = 3Extension: Point: y = 9 − 12 + 7 = 4, so (3, 4). Gradient: dy/dx = 2x − 4 = 2. y − 4 = 2(x − 3), so y = 2x − 2.
How do you classify a stationary point using the sign of dy/dx?Extension: Find dy/dx just before and just after. + then −: maximum. − then +: minimum. Same sign both sides: point of inflection (for example y = x³ at the origin).
Velocity–time graph: what does the area under it show?Extension: The distance travelled. Split the area into triangles, rectangles and trapezia; for a curved graph, estimate the area.

3.2.3 Solving equations and inequalities

Solve 2(3x − 1) − (x − 7) = 20: 6x − 2 − x + 7 = 20 (the minus changes both signs in the second bracket) 5x + 5 = 20, so x = 3
Solve (x + 3)/4 = (2x − 1)/3: Multiply both sides by 12: 3(x + 3) = 4(2x − 1) 3x + 9 = 8x − 4 13 = 5x, so x = 13/5 = 2.6
How do you solve f(x) = g(x) with a graph?: Draw y = f(x) and y = g(x). The solutions are the x-coordinates of the crossing points. A graph gives approximate values; algebra gives exact ones.
Three consecutive integers add up to 87. Find them.: x + (x + 1) + (x + 2) = 87 3x + 3 = 87, so x = 28. The integers are 28, 29 and 30.
Solve 2x + 3y = 12 and 3x − y = 7: Multiply the second by 3: 9x − 3y = 21. Add to the first: 11x = 33, so x = 3. Then 6 + 3y = 12, so y = 2. Check: 9 − 2 = 7.
Simultaneous linear equations: what if both letters vanish?: A false statement (for example 0 = 5): the lines are parallel, so there is no solution. A true statement (0 = 0): it is the same line, with infinitely many solutions.
Solve y = x² − 3x and y = x + 5Extension: x² − 3x = x + 5, so x² − 4x − 5 = 0 and (x − 5)(x + 1) = 0. Find y from the linear equation: (5, 10) and (−1, 4).
Solve 2x² − 8x = 0: Factorise: 2x(x − 4) = 0, so x = 0 or x = 4. Do not divide by x, or you lose x = 0.
Solve 3x² + 11x − 4 = 0Extension: ac = −12; 12 and −1 have product −12 and sum 11. 3x² + 12x − x − 4 = 3x(x + 4) − (x + 4) = (x + 4)(3x − 1) = 0 x = −4 or x = 1/3
Solve 2x² + 5x − 1 = 0, giving answers to 3 significant figuresExtension: x = (−b ± √(b² − 4ac))/(2a), with b² − 4ac = 25 + 8 = 33. x = (−5 ± √33)/4, so x = 0.186 or x = −2.69.
Solve x² − 6x + 2 = 0 by completing the squareExtension: (x − 3)² − 9 + 2 = 0, so (x − 3)² = 7. x = 3 ± √7 (5.65 or 0.35 to 2 d.p.)
Solve 6 − 4x ≥ 18: −4x ≥ 12. Dividing by a negative reverses the sign: x ≤ −3.
Number lines: open and closed circles: Open circle: < or >, the boundary is not included. Closed circle: ≤ or ≥, the boundary is included. The integers in −1 ≤ x < 4 are −1, 0, 1, 2, 3.
Solve x² + x − 12 > 0Extension: (x + 4)(x − 3) > 0; critical values −4 and 3. "> 0" means outside the roots: x < −4 or x > 3. ("< 0" would be between the roots.)
Inequalities on a graph: line style and which sideExtension: Solid line for ≤ or ≥; dashed line for < or >. Test a point not on the line, such as (0, 0): if it satisfies the inequality, that side is the region.
Turn into symbols: "at least", "at most", "from 3 to 8 inclusive": At least: ≥. At most: ≤. From 3 to 8 inclusive: 3 ≤ x ≤ 8.

3.2.4 Sequences

Term-to-term rule: A rule that builds each term from the previous one. It needs a starting term.
Position-to-term rule: A formula giving a term from its position, also called the nth term.
Start at 4 and use the rule "multiply by 3, then subtract 2". Write the first three terms.: 4, then 3 × 4 − 2 = 10, then 3 × 10 − 2 = 28. Terms: 4, 10, 28
Find the nth term of 7, 10, 13, 16, … and of 30, 25, 20, 15, …: d = 3, constant 7 − 3 = 4: 3n + 4. d = −5, constant 30 + 5 = 35: 35 − 5n.
The 7th term of an arithmetic sequence is 40 and the common difference is −3. Find the nth term.: nth term = −3n + c. Use the 7th term: −21 + c = 40, so c = 61. nth term = 61 − 3n (check: n = 7 gives 40).
Is 99 a term of the sequence with nth term 4n − 1? Is 60?: 4n − 1 = 99 gives n = 25, a whole number: yes, the 25th term. 4n − 1 = 60 gives n = 15.25: no.
Triangular numbers and their nth term: 1, 3, 6, 10, 15, 21, … nth term = n(n + 1)/2. The differences go up by 1 each time.
Quadratic sequence: how is the second difference linked to the nth term an² + bn + c?Extension: Second difference = 2a, so a = half the second difference. Then subtract an² from the sequence and find the linear nth term bn + c of what is left.
Find the nth term of 6, 15, 28, 45, 66, …Extension: First differences 9, 13, 17, 21; second difference 4, so a = 2. Sequence − 2n²: 4, 7, 10, 13, 16, which is 3n + 1. nth term = 2n² + 3n + 1