How to use this bank.
- Start at Level 1 and move up — don't jump to Level 4 first.
- For any surd equation, always check your answer in the ORIGINAL equation, not the squared version — some questions here are deliberately built to test this.
- For quadratic inequalities, decide whether the parabola opens up or down before reading off the answer.
- 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
24%
Level 1 | Knowledge
Direct Recall
One-step applications of a rule already stated — rational exponents, surds, the discriminant, and completing the square.
Rational Exponents
Square Root as a Rational Exponent
Evaluate: \(25^{\frac{1}{2}}\)
Memo
✓ In a rational exponent \(x^{p/q}\), the denominator \(q\) is the root being taken — here the denominator is \(2\), a square root.✓ \(25^{\frac12}=\sqrt{25}=\boxed{5}\)Surds
Simplify a Surd
Simplify: \(\sqrt{20}\)
Memo
✓ Using \(\sqrt{a\times b}=\sqrt a\times\sqrt b\) in reverse, split \(20\) into its largest perfect-square factor and whatever remains: \(\sqrt{20}=\sqrt{4\times5}\)✓ Take the square root of the perfect-square part only: \(\sqrt4\times\sqrt5=\boxed{2\sqrt5}\)Nature of Roots
Calculate the Discriminant
Calculate the discriminant of \(2x^2+3x+1=0\).
Memo
✓ Read off \(a=2,\,b=3,\,c=1\) directly from the standard form \(ax^2+bx+c=0\), then substitute into the discriminant formula: \(\Delta=b^2-4ac=3^2-4(2)(1)=9-8\)✓ \(\boxed{\Delta=1}\)Completing the Square
The Completing-the-Square Constant
What number must be added to \(x^2+10x\) to make it a perfect square trinomial?
Memo
✓ Completing the square always adds the SAME number: half the coefficient of \(x\), squared: \(\left(\dfrac{10}{2}\right)^2=5^2\)✓ \(\boxed{25}\) — this makes \(x^2+10x+25=(x+5)^2\), a genuine perfect square29%
Level 2 | Routine Procedures
One Established Method
Rational exponents, surd arithmetic, a simple surd equation, completing the square, and the quadratic formula, each applied cleanly.
Rational Exponents
Cube Root Raised to a Power
Evaluate: \(8^{\frac{2}{3}}\)
Memo
✓ In \(8^{\frac23}\), the denominator \(3\) is the root and the numerator \(2\) is the power — take the root first, since it keeps the numbers small: \(\sqrt[3]{8}=2\)✓ Raise that result to the power given by the numerator: \(\boxed{2^2=4}\)Surds
Combining Like Surds
Simplify: \(4\sqrt5-\sqrt5+2\sqrt5\)
Memo
✓ All three terms already share \(\sqrt5\), so only the coefficients in front combine, exactly like combining \(4x-x+2x\): \(4-1+2=5\)✓ \(\boxed{5\sqrt5}\)Surd Equations
A Basic Surd Equation
Solve for \(x\): \(\sqrt{x+2}=3\)
Memo
✓ The surd is already isolated, so square both sides to remove the root: \(x+2=9\)✓ \(x=7\)✓ Squaring can introduce a false solution, so this must be checked in the ORIGINAL equation: \(\sqrt{7+2}=\sqrt9=3\)✓ valid — \(\boxed{x=7}\)Completing the Square
Solve by Completing the Square
Solve for \(x\) by completing the square: \(x^2+4x-5=0\)
Memo
✓ Move the constant to the other side so only the \(x\)-terms remain on the left: \(x^2+4x=5\)✓ Add \(\left(\dfrac{4}{2}\right)^2=4\) to BOTH sides (the amount that turns the left side into a perfect square): \((x+2)^2=9\)✓ Take the square root of both sides — never forget the \(\pm\), or one whole solution is lost: \(x+2=\pm3\)✓ \(\boxed{x=1\text{ or }x=-5}\)Quadratic Formula
Solve Using the Quadratic Formula
Solve for \(x\) using the quadratic formula: \(2x^2+5x-3=0\)
Memo
✓ Identify \(a=2,\,b=5,\,c=-3\) and compute the discriminant first, since it appears under the square root: \(\Delta=25-4(2)(-3)=25+24=49\)✓ Substitute into the quadratic formula: \(x=\dfrac{-5\pm\sqrt{49}}{4}=\dfrac{-5\pm7}{4}\)✓ Evaluate both signs separately: \(\boxed{x=0{,}5\text{ or }x=-3}\)24%
Level 3 | Complex Procedures
Multi-Step Methods
Nature of roots, a surd equation with an extraneous root, an irrational-root quadratic, and a basic quadratic inequality.
Nature of Roots
No Real Roots
Determine the nature of the roots of \(3x^2-2x+1=0\), without solving the equation.
Memo
✓ The question only asks for the NATURE of the roots, so there is no need to solve the equation — only the discriminant is needed: \(a=3,\,b=-2,\,c=1\): \(\Delta=(-2)^2-4(3)(1)=4-12=-8\)✓ \(\Delta<0\) means the quadratic formula would need the square root of a negative number, which has no real value: \(\boxed{\text{there are no real roots}}\)Surd Equations
A Surd Equation With an Extraneous Root
Solve for \(x\): \(\sqrt{2x+3}=x\)
Memo
✓ Square both sides to remove the root, then rearrange to standard form: \(2x+3=x^2\Rightarrow x^2-2x-3=0\)✓ Factorise and apply the zero-product law: \((x-3)(x+1)=0\Rightarrow x=3\) or \(x=-1\)✓ Squaring is not reversible, so every root must be checked in the ORIGINAL (unsquared) equation. Check \(x=3\): \(\sqrt9=3=3\)✓ valid✓ Check \(x=-1\): \(\sqrt1=1\), but the original equation needs this to equal \(x=-1\), and \(1\neq-1\)✗ — reject this extraneous root (a square root can never equal a negative number)✓ \(\boxed{x=3}\) onlyQuadratic Formula
Irrational Roots
Solve for \(x\) using the quadratic formula, correct to two decimal places: \(x^2-4x-1=0\)
Memo
✓ Identify \(a=1,\,b=-4,\,c=-1\) and compute the discriminant: \(\Delta=16+4=20\) — not a perfect square, confirming the roots are irrational as the question implies✓ Substitute into the formula, keeping the whole numerator over \(2a\): \(x=\dfrac{4\pm\sqrt{20}}{2}=2\pm\sqrt5\)✓ Evaluate to two decimal places: \(\boxed{x\approx4{,}24\text{ or }x\approx-0{,}24}\)Quadratic Inequalities
A Basic Quadratic Inequality
Solve for \(x\): \(x^2-5x+6\le0\)
Memo
✓ Solve the associated equation to find the critical values first: \(x^2-5x+6=0\Rightarrow(x-2)(x-3)=0\Rightarrow x=2\) or \(x=3\)✓ The leading coefficient is positive, so this parabola opens upward — it is negative (\(\le0\)) between its two roots, not outside them✓ Since the inequality is \(\le\), not \(<\), the endpoints are included: \(\boxed{2\le x\le3}\)23%
Level 4 | Problem Solving
Combined Skills
Simultaneous linear-quadratic equations, a downward-opening quadratic inequality, solving for an unknown coefficient, and a word problem.
Simultaneous Equations
Simultaneous Linear and Quadratic Equations
Solve simultaneously: \(y=2x-1\) and \(y=x^2-4\)
Memo
✓ Both equations already equal \(y\), so set them equal to each other: \(2x-1=x^2-4\)✓ Rearrange to standard form and solve: \(x^2-2x-3=0\Rightarrow(x-3)(x+1)=0\Rightarrow x=3\) or \(x=-1\)✓ Substitute each \(x\)-value into the LINEAR equation (simpler than the quadratic) to find the matching \(y\): \(x=3\Rightarrow y=5\); \(x=-1\Rightarrow y=-3\)✓ \(\boxed{(3,5)\text{ and }(-1,-3)}\) — a line and a parabola can intersect at most twice, matching the two pairs found hereQuadratic Inequalities
A Downward-Opening Quadratic Inequality
Solve for \(x\): \(-x^2+x+6\ge0\)
Memo
✓ The leading coefficient is negative, which makes the shape harder to reason about directly — multiply both sides by \(-1\) to flip it into the familiar upward case, and FLIP the inequality sign since the multiplier is negative: \(x^2-x-6\le0\)✓ Solve the associated equation for the critical values: \((x-3)(x+2)=0\Rightarrow x=3\) or \(x=-2\)✓ This rearranged parabola now opens upward, so it is \(\le0\) between its roots (inclusive)✓ \(\boxed{-2\le x\le3}\) — check with a value inside the interval, \(x=0\), in the ORIGINAL inequality: \(-0+0+6=6\ge0\)✓Nature of Roots
Solving for an Unknown Coefficient
Determine the value of \(k>0\) for which \(x^2+kx+9=0\) has equal roots.
Memo
✓ "Equal roots" is exactly the CAPS-named discriminant condition \(\Delta=0\) — set it up and substitute \(a=1,\,b=k,\,c=9\): \(k^2-4(1)(9)=0\)✓ Solve this equation for \(k\): \(k^2=36\Rightarrow k=\pm6\)✓ The question restricts to \(k>0\), so reject the negative root: \(\boxed{k=6}\)Word ProblemQuadratic Formula
A Number and Its Reciprocal
A positive number, added to its own reciprocal, gives \(2{,}5\). Determine the number(s).
Memo
✓ Let the number be \(x\); "added to its own reciprocal" translates directly: \(x+\dfrac{1}{x}=2{,}5\)✓ Multiply through by \(x\) to clear the fraction and turn this into a quadratic: \(x^2+1=2{,}5x\Rightarrow x^2-2{,}5x+1=0\)✓ Multiply every term by \(2\) to clear the decimal coefficient before factorising: \(2x^2-5x+2=0\Rightarrow(2x-1)(x-2)=0\)✓ \(\boxed{x=0{,}5\text{ or }x=2}\) — check both in the original equation: \(0{,}5+2=2{,}5\)✓ and \(2+0{,}5=2{,}5\)✓, and both are positive as required