Flashcards · 8300 · Topic 2

Algebra flashcards

42 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, vocabulary and manipulation

Term (in algebra): One part of an expression, such as −5x in 2x² − 5x.
Equation compared with identity: An equation (=) is true for some values; an identity (≡) is true for all values.
Expand compared with factorise: Expanding removes brackets; factorising puts them in.
Difference of two squares: a² − b² = (a + b)(a − b). It has no middle term. Example: p² − 64 = (p + 8)(p − 8).
How do you factorise x² + bx + c?: 1. List the factor pairs of c. 2. Pick the pair that adds to b. 3. Write (x + p)(x + q) and expand to check. Example: x² + 9x + 20 = (x + 4)(x + 5).
How do you change the subject of a formula?: 1. Clear fractions and expand brackets. 2. Collect the terms with the new subject on one side. 3. If it appears in two terms, factorise it out. 4. Divide, then take a root if needed. Example: v² = u² + 2as → u = ±√(v² − 2as).
How do you factorise ax² + bx + c (a ≠ 1)?Higher: 1. Work out a × c. 2. Find two numbers with product ac and sum b. 3. Split bx into those two terms and factorise in pairs. Example: 2x² − 7x + 3 = 2x(x − 3) − 1(x − 3) = (2x − 1)(x − 3).
fg(x) compared with gf(x)Higher: The inner function acts first: fg(x) means g first, then f. They are usually different.
f⁻¹(x) compared with 1/f(x)Higher: The inverse function is not the reciprocal.
How do you find an inverse function f⁻¹(x)?Higher: 1. Write y = f(x). 2. Make x the subject. 3. Swap back to x. Example: f(x) = 2x + 7 gives f⁻¹(x) = (x − 7)/2.

3.2.2 Graphs

Gradient of the line through (x₁, y₁) and (x₂, y₂): m = (y₂ − y₁) ÷ (x₂ − x₁), subtracting in the same order top and bottom. Example: (2, 5) and (6, 17): 12 ÷ 4 = 3.
In y = mx + c, what are m and c?: m is the gradient; c is the y-intercept. Rearrange to y = … first: 3x + 2y = 12 → y = −1.5x + 6, gradient −3/2, intercept 6.
Gradients of perpendicular linesHigher: m₁ × m₂ = −1, so m₂ = −1/m₁. A line perpendicular to y = 4x − 7 has gradient −1/4.
Turning point of y = (x + p)² + qHigher: (−p, q). Example: x² + 10x + 18 = (x + 5)² − 7, turning point (−5, −7).
Shapes of y = x³ and y = 1/x: y = x³: an S-shape through the origin, flat at (0, 0). y = 1/x: two branches in the first and third quadrants that never touch the axes.
Equation of a circle centre the origin, and its tangentHigher: x² + y² = r² has centre (0, 0) and radius r. A tangent is perpendicular to the radius at that point, so its gradient is the negative reciprocal of the radius gradient.
Effect on the point (a, b): y = f(x) + k and y = f(x + k)Higher: y = f(x) + k: (a, b + k), a move up by k. y = f(x + k): (a − k, b), a move left by k (opposite to the sign).
Effect on the point (a, b): y = −f(x) and y = f(−x)Higher: y = −f(x): (a, −b), a reflection in the x-axis. y = f(−x): (−a, b), a reflection in the y-axis.
Distance-time and velocity-time graphs: what does the gradient of a straight section give?: Distance-time: gradient = speed; a horizontal section means stationary. Velocity-time: gradient = acceleration; a horizontal section means constant velocity.
How do you estimate the area under a curve?Higher: Split it into equal-width trapezia: area ≈ (h/2) × [first + 2(middle values) + last]. Say whether it is an over- or underestimate: if the curve bends down, the trapezia sit below it (underestimate).
What does the area under a velocity-time graph give?Higher: The distance travelled. Split the area into triangles, rectangles and trapezia; for a curve, estimate it with trapezia.
Gradients of parallel lines: Parallel lines have the same gradient. y = 3x + 1 and y = 3x − 4 are parallel (both have gradient 3).

3.2.3 Solving equations and inequalities

How do you solve a quadratic equation by factorising?: 1. Rearrange to … = 0. 2. Factorise. 3. Set each bracket equal to 0. Example: x² − 7x + 12 = 0 → (x − 3)(x − 4) = 0 → x = 3 or x = 4.
Why not divide x² − 8x = 0 by x?: You lose the solution x = 0. Factorise instead: x(x − 8) = 0, so x = 0 or x = 8.
Quadratic formulaHigher: For ax² + bx + c = 0 (a ≠ 0): x = (−b ± √(b² − 4ac)) / (2a) It is not given in the exam.
Solve x² + 4x − 1 = 0 by completing the squareHigher: (x + 2)² − 4 − 1 = 0, so (x + 2)² = 5 x + 2 = ±√5 x = −2 ± √5
How do you solve linear simultaneous equations by elimination?: 1. Line up the x and y terms. 2. Multiply to make one pair of coefficients equal. 3. Same signs: subtract; different signs: add. 4. Solve, substitute back, check. Example: 2x − y = 7, x + y = 8 → 3x = 15, x = 5, y = 3.
How do you solve one linear and one quadratic equation simultaneously?Higher: 1. Make y (or x) the subject of the linear equation. 2. Substitute into the quadratic. 3. Rearrange to = 0 and solve. 4. Find the matching value of the other letter for each solution.
When does an inequality sign reverse?: Only when you multiply or divide both sides by a negative number. Example: 7 − 3x ≤ 1 → −3x ≤ −6 → x ≥ 2.
Number line: open circle or closed circle?: Open circle for < or > (the end value is not included). Closed circle for ≤ or ≥ (it is included).
How do you solve a quadratic inequality such as x² − 4x − 5 < 0?Higher: Factorise to find the critical values: (x − 5)(x + 1), so −1 and 5. Sketch the U-shaped curve. < 0: between the roots, −1 < x < 5. > 0 would be outside: x < −1 or x > 5.
Iteration: what do xₙ and xₙ₊₁ mean, and how do you show a root lies between a and b?Higher: xₙ is the current value and xₙ₊₁ = g(xₙ) is the next. Feed each answer back in and keep full values. A root of f(x) = 0 lies between a and b if f(a) and f(b) have opposite signs.

3.2.4 Sequences

Term-to-term rule compared with position-to-term rule: Term-to-term: how to get the next term from the previous one (needs a first term). Position-to-term (nth term): a formula in n that gives any term directly.
How do you find the nth term of a linear sequence?: 1. The common difference d gives dn. 2. Compare dn with the sequence and add or subtract the gap. 3. Check one term. Example: 40, 37, 34, 31 → d = −3; nth term 43 − 3n.
Is 75 a term of the sequence 4n − 1?: Set 4n − 1 = 75, so 4n = 76 and n = 19. n is a positive whole number, so yes: the 19th term.
Triangular numbers: 1, 3, 6, 10, 15, 21, … (add 2, 3, 4, 5, …). nth term n(n + 1)/2.
Fibonacci-type sequence: Each term is the sum of the two before it. Starting a, b: a, b, a + b, a + 2b, 2a + 3b, …
Geometric progression: how do you find the common ratio r?: r = any term ÷ the term before it (check with a second pair). Example: 54, 18, 6, … has r = 18 ÷ 54 = 1/3, so the next term is 2.
How do you recognise a quadratic sequence?: The second differences (differences between consecutive first differences) are constant.
How do you find the nth term of a quadratic sequence?Higher: 1. a = second difference ÷ 2; start with an². 2. Subtract an² from every term. 3. Find the nth term bn + c of what remains. 4. nth term = an² + bn + c; check one term.
Find the nth term of 2, 5, 9, 14, 20, …Higher: Second difference 1, so a = 1/2. Term − ½n²: 1.5, 3, 4.5, 6, 7.5 (d = 1.5, gap 0). nth term = ½n² + (3/2)n. Check n = 3: 4.5 + 4.5 = 9.
Geometric sequence with first term 4 and common ratio √5Higher: 4, 4√5, 20, 20√5, 100, … Keep surds exact: √5 × √5 = 5, so every second term is multiplied by 5.