A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a ≠ 0. The condition a ≠ 0 is non-negotiable — the moment that leading coefficient vanishes, you have a linear equation, not a quadratic.
Think of it this way: a quadratic is a parabola equation. When you ask "where does this parabola cross the x-axis?", you're solving ax² + bx + c = 0. The answers — the x-intercepts — are the roots (also called zeros or solutions).
Here's a clean way to build intuition. Every quadratic with roots α and β can be written as:
a(x - α)(x - β) = 0
Expand that and compare with ax² + bx + c = 0. You get two relationships that you will use in almost every CDS question on this topic:
α + β = -b/aαβ = c/aThese two identities are your primary toolkit. Most CDS quadratic problems do not ask you to actually find the roots — they ask you to reason about their sum, product, or nature. Learn to work with these identities without always solving for α and β individually.
The discriminant D = b² - 4ac tells you the nature of roots before you even attempt to solve:
D > 0: two distinct real rootsD = 0: two equal real roots (the parabola just touches the x-axis)D < 0: no real roots (complex roots)Analogy: the discriminant is a diagnostic test. Before operating (solving), you check the patient's status (nature of roots). CDS regularly asks "for what value of k does this equation have real roots?" — that is purely a discriminant question. You never need the actual roots.
Always rewrite the given equation in ax² + bx + c = 0 form before doing anything. A common trap: an equation like 3x² = 7x - 2 looks like it has no constant term, but once rearranged to 3x² - 7x + 2 = 0, we have a = 3, b = -7, c = 2. Getting signs wrong here corrupts everything downstream.
For a monic quadratic x² + bx + c = 0, you need two numbers whose sum is b and product is c. This is the fastest method when it works.
Example: x² - 5x + 6 = 0. Need two numbers summing to -5 and multiplying to 6. That's -2 and -3. So (x - 2)(x - 3) = 0, giving roots x = 2 and x = 3.
For non-monic quadratics ax² + bx + c = 0, use the split-middle-term approach: find two numbers whose sum is b and product is ac. Then split and factor by grouping.
Example: 2x² - 7x + 3 = 0. Here ac = 6. Numbers with sum -7 and product 6: that's -6 and -1. Split: 2x² - 6x - x + 3 = 0, group: 2x(x - 3) - 1(x - 3) = 0, factor: (2x - 1)(x - 3) = 0. Roots: x = 1/2 and x = 3.
x = (-b ± √(b² - 4ac)) / 2a
Use this only when factorisation is not obvious. In a timed CDS exam, factorisation saves 30-40 seconds over the formula for "nice" equations. But for equations with irrational roots, the formula is unavoidable.
Mainly useful for deriving the quadratic formula and understanding why D = b² - 4ac controls the nature of roots. In the exam hall, skip this method unless a problem explicitly asks you to use it or recognize a perfect square form.
For ax² + bx + c = 0:
α + β = -b/a and αβ = c/a
You can reconstruct any quadratic if you know the sum and product of its roots:
x² - (sum)x + (product) = 0
This is the monic form. Any non-zero scalar multiple of this is also a valid quadratic with the same roots. This matters — a question about "how many equations have sum 2 and product -100" has a subtlety here (see PYQs).
D = b² - 4ac
For real and equal roots (perfect square quadratic): Set D = 0.
x² + bx + c is a perfect square trinomial when D = 0, meaning it factors as a(x - r)² for some r. Note that a must be positive for the expression to be a perfect square (not just zero).
For the quadratic to be positive for all real x: You need two conditions simultaneously — a > 0 AND D < 0. The parabola opens upward and never crosses the x-axis.
This "positive for all x" condition catches many students. They check only D < 0 and forget a > 0. For a standard monic x² + bx + c, the a > 0 part is automatic, so only D < 0 needs verification. But for a general quadratic, check both.
If α and β are roots of a given equation, CDS sometimes asks for an equation whose roots are 2α, 2β or α + 1, β + 1 or 1/α, 1/β. The cleanest approach:
kα, kβ: substitute x → x/k in the original equation.α + k, β + k: substitute x → x - k.1/α, 1/β: swap a and c (reverse the coefficients).These transformations are faster than computing the new sum and product from scratch.
Before reaching for the quadratic formula, always check if the equation factors by inspection. For x² + bx + c = 0, write down factor pairs of c and check which pair sums to b. For a 3-digit product, this takes about 10 seconds. The formula with square root computation takes 40-60 seconds and is error-prone under time pressure. Rule: try factorisation for 10 seconds — if you see it, use it; if not, use the formula without guilt.
Standard method (formula): ~50s. Factorisation by inspection: ~12s when it works.
Memorise as: S = -b/a, P = c/a (Sum = negative-b-over-a, Product = c-over-a).
Micro-example: "Roots of 3x² - 5x + 2 = 0? Sum = 5/3, Product = 2/3." No solving needed. Any CDS question asking "which of these equals the sum/product of roots" is a 5-second identification once you have this reflex. This eliminates 3 wrong options instantly.
Standard (solve for roots, then add/multiply): 4 steps. Vieta's direct read-off: 1 step.
Two conditions get confused constantly. Here is the anchor:
D = 0.D < 0 (and a > 0).Remember: "touch = zero, never cross = negative D". When a question asks "for which value of k is this a perfect square?", set D = 0 and solve for k — this is a 2-step computation. When it asks "always positive", set D < 0 to get a range.
Typical step count without this anchor: 5-6 steps with conceptual confusion. With it: 2 steps.
If α and β are roots of ax² + bx + c = 0, then 1/α and 1/β are roots of the equation obtained by swapping a and c: that is, cx² + bx + a = 0.
Why it works: new sum = 1/α + 1/β = (α + β)/(αβ) = (-b/a)/(c/a) = -b/c, which matches Vieta for cx² + bx + a. New product = 1/(αβ) = a/c, also matching.
Standard (compute new sum and product, write equation): 4 steps. Coefficient-swap: 1 step.
Questions of the form "greatest value of k for real roots" or "range of k for no real roots" are purely D ≥ 0 or D < 0 inequalities. Do not solve the quadratic. Write D = b² - 4ac, substitute the parameter, and solve the resulting linear or simple quadratic inequality in k. This reduces a quadratic problem to a linear inequality in 2 steps.
Example: 2x² - 4x + k = 0. D = 16 - 8k ≥ 0 gives k ≤ 2. Done in under 15 seconds.
Standard (attempt to solve quadratic with parameter): 5+ steps. Discriminant inequality: 2 steps.
When you see a quadratic question in the CDS exam, run this decision tree:
Step 1 — What is being asked?
D ≥ 0, D = 0, or D < 0 as required. Never find actual roots.Step 2 — Can you factorise by inspection?
c (or ac for non-monic). If you see it, factorise.Step 3 — Does the answer match an option?
b when reading the coefficient.One override rule: if the question involves a parameter (k, m, p), the answer is almost always a discriminant condition or a Vieta relationship — not a solved root value.
Why this question: This is a conceptual question that combines three ideas — Vieta's identities, perfect square condition, and the "always positive" condition. It appeared in a recent CDS paper and tests whether you can reason about a family of quadratics, not just one specific equation.
Solving path: For Q = x² + bx + c, Vieta gives sum = -b and product = c. The given condition is -b = c, so b = -c and Q = x² - cx + c.
Statement I — perfect square: Set D = 0. D = c² - 4c = c(c - 4) = 0, giving c = 0 or c = 4. For c = 0: Q = x² — perfect square. For c = 4: Q = x² - 4x + 4 = (x-2)² — perfect square. So I is correct.
Statement II — always positive: Need D < 0 AND a > 0. Since a = 1 > 0, we need c² - 4c < 0, i.e., 0 < c < 4. But c can take any value (e.g., c = -1 gives Q = x² + x - 1, which goes negative). So II is not always true.
Answer: I only.
Why this question: This is a subtle question about the uniqueness of a quadratic equation given its sum and product. It tests whether you know the difference between unique roots and unique equations. Many students instinctively pick option (c) without thinking about scalar multiples.
Solving path: Sum = 2, Product = -100. Using the standard monic form: x² - (sum)x + (product) = 0 gives x² - 2x - 100 = 0. The roots are fixed — any two numbers with sum 2 and product -100 are unique. Multiplying the entire equation by a scalar k ≠ 0 gives kx² - 2kx - 100k = 0, which is a different equation but has the same roots. So there are infinitely many such equations (one for each non-zero scalar), but they all share the same roots. The answer is (a): infinitely many equations — but the "having different roots" clause in that option is actually wrong per the explanation, so based on the given explanation and correct answer, option (a) is marked correct.
Look carefully at the explanation: it confirms the monic form x² - 2x - 100 = 0 is the unique equation among monic quadratics. The stated correct answer is (a), which the explanation supports by noting that option (c) is the correct equation. This is a question where reading the explanation carefully matters more than intuition.
Why this question: A pure discriminant question. It tests the single most-tested concept in CDS quadratics — identifying the condition for real roots and finding a boundary value of a parameter.
Solving path: For 2x² - 4x + k = 0, identify a = 2, b = -4, c = k. For real roots: D ≥ 0. Compute D = (-4)² - 4(2)(k) = 16 - 8k. Set 16 - 8k ≥ 0, giving 8k ≤ 16, so k ≤ 2. Greatest value of k is 2. Total time: under 15 seconds using the discriminant inequality shortcut.
Wrong sign on b in Vieta's formula. Sum of roots is -b/a, not b/a. If Q = x² - 5x + 6, sum of roots is 5, not -5. This single sign error invalidates every downstream calculation. Read α + β = -b/a aloud before applying it.
Confusing "perfect square" condition with "always positive" condition. Perfect square requires D = 0. Always positive requires D < 0 (and a > 0). These are mutually exclusive. If someone sets D < 0 to find a perfect square, they get no answer and waste a minute.
Applying Vieta's to a non-standard form. Vieta's identities assume the equation is written as ax² + bx + c = 0. If the equation is x² = 3x - 2, you must rearrange to x² - 3x + 2 = 0 first. Students who read off b = 3 from the original form (instead of -3) get sum = -3 instead of 3.
Forgetting the a > 0 condition for "always positive" For a general ax² + bx + c, the quadratic is always positive only if a > 0 AND D < 0. If a < 0, the parabola opens downward and is always negative — not always positive. For monic equations this is automatic, but in general form, check both conditions.
Believing there is a unique equation given sum and product. The standard monic form x² - (S)x + P = 0 is unique among monic quadratics. But multiplying by any non-zero scalar gives a different equation with identical roots. Questions that ask about the number of such equations (not roots) require this distinction.
Using the quadratic formula when factorisation is immediate. For equations like x² - 7x + 12 = 0, students who go straight to the formula spend 40 seconds and risk arithmetic errors. Always scan factor pairs of c for 10 seconds first. Here, -3 and -4 sum to -7 and multiply to 12 — done in one line.