How to use this bank.
- Start at Level 1 and move up — don't jump to Level 4 first.
- Rearrange every equation to standard form before choosing a method.
- Check every root — substitute back into the original equation, not the rearranged one.
- Reveal the memo only after a genuine attempt.
L1 — Knowledge 4 Qs
L2 — Routine Procedures 5 Qs
L3 — Complex Procedures 4 Qs
L4 — Problem Solving 4 Qs
20%
Level 1 | Knowledge
Recall & Direct Application
One law or fact, one step. If these feel shaky, go back to the Summary Notes diagnostic before continuing.
Exponents
Zero Exponent Law
Simplify: \(5^0+2^3\)
Memo
✓ The zero-exponent law makes any nonzero base raised to the power 0 equal to 1: \(5^0=1\)✓ A cubed term is that base multiplied by itself three times: \(2^3=8\)✓ Add the two simplified terms: \(1+8=9\)Factorisation
Difference of Two Squares
Factorise: \(x^2-16\)
Memo
✓ Recognise \(x^2-16\) as a difference of two squares, since \(16=4^2\): \(x^2-16=x^2-4^2\)✓ Apply \(a^2-b^2=(a-b)(a+b)\) with \(a=x,\,b=4\): \(x^2-16=x^2-4^2=(x-4)(x+4)\)Exponents
Same-Base Exponential Equation
Determine the value of \(x\) if \(2^x=16\).
Memo
✓ To equate exponents, both sides need the same base — rewrite \(16\) as a power of \(2\): \(16=2^4\)✓ Now \(2^x=2^4\) share a base, so the exponents themselves must be equal: \(x=4\)Quadratic equations
Both Roots of a Simple Square
Solve for \(x\): \(x^2=25\)
Memo
✓ Taking the square root of both sides introduces a \(\pm\), since both a positive and a negative number square to give the same result: \(x=\pm\sqrt{25}\)✓ \(x=5\) or \(x=-5\) — both roots must be given, not just the positive one (there is no linear term here to break the symmetry)35%
Level 2 | Routine Procedures
Standard Methods, Clean Numbers
The bulk of Paper 1 Question 1–2 marks live here. Fluency matters more than speed at this stage.
Quadratic equations
Expand Before You Factorise
Solve for \(x\): \(x(x+3)=28\)
Memo
✓ Expand the bracket first, then move everything to one side so the equation equals zero — the variable is on both sides of a product, so it cannot be solved directly yet: \(x^2+3x=28\Rightarrow x^2+3x-28=0\)✓ Find two numbers that multiply to \(-28\) and add to \(3\): these are \(7\) and \(-4\), giving \((x+7)(x-4)=0\)✓ Zero-product law — set each factor to zero in turn: \(x=-7\) or \(x=4\)Quadratic equations
Factorising with a Leading Coefficient
Solve for \(x\): \(3x^2+5x-2=0\)
Memo
✓ Multiply the leading coefficient by the constant: \(3\times(-2)=-6\); find two numbers multiplying to \(-6\) and adding to \(5\): these are \(6\) and \(-1\), giving the factorisation \((3x-1)(x+2)=0\)✓ Zero-product law: \(x=\tfrac13\) or \(x=-2\)Surds
Solving a Surd Equation
Solve for \(x\): \(\sqrt{3x+1}=4\)
Memo
✓ The surd is already isolated, so square both sides to remove it: \(3x+1=16\)✓ Solve the resulting linear equation: \(x=5\). Squaring can introduce a false root, so this must be checked in the ORIGINAL equation: \(\sqrt{16}=4\) ✓Exponents
Exponential Equation, Shifted Exponent
Solve for \(x\): \(2^{x-3}=16\)
Memo
✓ Write \(16\) as a power of the same base, \(2\): \(16=2^4\)✓ With matching bases, equate the exponents and solve for \(x\): \(x-3=4\Rightarrow x=7\)Quadratic equations
Perfect-Square Trinomial (Equal Roots)
Solve for \(x\): \(x^2-6x+9=0\)
Memo
✓ Recognise this as a perfect square trinomial: two numbers multiplying to \(9\) and adding to \(-6\) are both \(-3\), giving \((x-3)^2=0\)✓ Since the repeated factor gives only ONE distinct value, this is exactly what an equal-roots case looks like algebraically: \(x=3\) (equal roots — one repeated solution)30%
Level 3 | Complex Procedures
Multi-Step Methods
The quadratic formula, inequalities, simultaneous equations and discriminant reasoning — no single method solves these in one line.
Quadratic formula
Expand, Rearrange, Then the Formula
Solve for \(x\), correct to two decimal places: \((3x-1)(x-4)=16\)
Memo
✓ Expand the bracket product first, then move everything to one side — this equation does not factorise cleanly, so the full expand-then-solve route is needed: \(3x^2-13x+4=16\Rightarrow3x^2-13x-12=0\)✓ Read off \(a=3,b=-13,c=-12\)✓ Compute the discriminant: \(\Delta=169+144=313\) — not a perfect square, matching the instruction to answer correct to two decimal places✓ Substitute into the quadratic formula, keeping the whole numerator over \(2a\): \(x=\dfrac{13\pm\sqrt{313}}{6}\)✓ Evaluate both signs to two decimal places: \(x\approx5{,}12\) or \(x\approx-0{,}78\)Quadratic inequalities
Quadratic Inequality (Strict)
Solve for \(x\): \(x^2-x-12>0\)
Memo
✓ Solve the associated equation first to find the critical values: \((x-4)(x+3)=0\Rightarrow x=4\) or \(x=-3\)✓ The leading coefficient is positive, so this parabola opens upward — it is positive (\(>0\)) OUTSIDE its two roots, not between them✓ The inequality is strict (\(>\), not \(\ge\)), so the endpoints themselves are excluded: \(x<-3\) or \(x>4\)Simultaneous equations
Line and Parabola
Solve simultaneously for \(x\) and \(y\): \(y=x-2\) and \(y=x^2-4x+2\)
Memo
✓ Both equations already equal \(y\), so set them equal to each other and rearrange to standard form: \(x-2=x^2-4x+2\Rightarrow x^2-5x+4=0\)✓ Factorise and apply the zero-product law: \((x-4)(x-1)=0\Rightarrow x=4\) or \(x=1\)✓ Substitute each \(x\)-value into the LINEAR equation (simpler than the quadratic) to find the matching \(y\): using \(y=x-2\): \((4;2)\) and \((1;-1)\)Nature of roots
Finding k for Real Roots
For which value(s) of \(k\) will \(x^2-4x+k=0\) have real roots?
Memo
✓ "Real roots" covers both the equal AND unequal cases, so the condition is \(\Delta\geq0\), not a strict inequality✓ Substitute \(a=1,b=-4,c=k\) into the discriminant and set up the inequality: \(16-4k\geq0\)✓ Solve this inequality for \(k\) — dividing by a positive number never flips the sign: \(k\leq4\)15%
Level 4 | Problem Solving
Unfamiliar Context, Full Riders
Translate words into equations first — the algebra itself is usually routine once the equation is set up correctly.
Word problem
Consecutive Integers
The product of two consecutive positive integers is 132. Determine the two integers.
Memo
✓ Let the smaller integer be \(x\); "consecutive" makes the next one \(x+1\)✓ Translate "product is 132" directly into an equation, then move everything to one side: \(x(x+1)=132\Rightarrow x^2+x-132=0\)✓ \(529\) happens to be a perfect square, \(23^2\), so the quadratic formula gives exact integer roots: \(\Delta=1+528=529=23^2\Rightarrow x=\dfrac{-1\pm23}{2}\)✓ The problem specifies positive integers, so reject the negative solution: \(x=11\) (reject \(x=-12\)); the integers are 11 and 12Word problem
Projectile Height
A ball's height above the ground, \(t\) seconds after being thrown, is given by \(h=20t-5t^2\) (in metres). Determine the time(s) at which the ball is at ground level.
Memo
✓ "Ground level" means the height is zero: \(h=0\)✓ Both terms in \(20t-5t^2=0\) share a common factor of \(5t\); factorising instead of dividing by \(t\) keeps both solutions visible: \(20t-5t^2=0\Rightarrow5t(4-t)=0\)✓ Zero-product law gives two times: \(t=0\) (the instant the ball is thrown) or \(t=4\) (when it lands) — the physically meaningful answer to when it returns to ground level is \(t=4\) secondsSimultaneous equations
Sum and Product of Two Numbers
Two numbers \(x\) and \(y\) satisfy \(x+y=7\) and \(xy=12\). Solve for \(x\) and \(y\) simultaneously.
Memo
✓ Rearrange the sum equation to make \(y\) the subject: \(y=7-x\)✓ Substitute into the product equation and rearrange to standard form: \(x(7-x)=12\Rightarrow x^2-7x+12=0\Rightarrow(x-3)(x-4)=0\)✓ Each \(x\)-value pairs with its matching \(y\) from \(y=7-x\): \(x=3,y=4\) or \(x=4,y=3\) — both orderings are genuinely valid, since the question never states which of \(x,y\) is largerNature of rootsQuadratic formula
Nature of Roots, then Solve
(1) Show that the roots of \(x^2-2x-2=0\) are real, irrational and unequal. (3)
(2) Hence solve for \(x\), correct to two decimal places. (3)
Memo
✓ Part (1) only needs the discriminant, not a full solve: \(a=1,b=-2,c=-2\): \(\Delta=(-2)^2-4(1)(-2)=4+8=12\)✓ \(\Delta>0\) confirms two distinct real roots; since \(12\) is not a perfect square, those roots are irrational — exactly the three properties the question asks to show✓ Part (2) reuses the same \(\Delta\) already found — substitute into the quadratic formula: \(x=\dfrac{2\pm\sqrt{12}}{2}=1\pm\sqrt3\)✓ Evaluate to two decimal places: \(x\approx2{,}73\) or \(x\approx-0{,}73\)