Quadratic Equations and Inequalities
~14 min read
- What: Quadratic equations (degree 2 polynomials) and algebraic inequalities tested directly in NDA Paper I Maths, spanning roots, discriminant, Vieta's formulas, and solution sets.
- Why it matters: This topic has appeared in every NDA paper from 2010 to 2023, often contributing 4–8 questions per sitting — one of the most reliable scoring zones.
- Key fact: Roughly 70 % of questions test either Vieta's sum/product relations or the nature-of-roots condition (discriminant ≥ 0), making those two tools your highest-ROI preparation targets.
Quadratic equations sit at the intersection of algebra and real analysis in the NDA syllabus. You need them to solve polynomial problems, understand complex roots, handle modulus-based equations, and work with inequalities. If you can crack discriminant conditions and Vieta's formulas cold, you will pick up marks quickly — most questions resolve in under two minutes once you see the pattern.
What This Topic Covers
The NDA Maths syllabus groups Quadratic Equations and Inequalities under the broader Algebra section. You are expected to handle the following:
- Standard form \(ax^2 + bx + c = 0\) and finding roots by factoring, completing the square, or the quadratic formula.
- Nature of roots using the discriminant \(D = b^2 - 4ac\).
- Sum and product of roots (Vieta's formulas).
- Forming a new quadratic given its roots or given transformed roots (e.g., roots squared, reciprocal roots).
- Equations reducible to quadratic form (e.g., biquadratic, exponential substitutions).
- Modulus equations: \(|x^2 - 3x + 2| > x^2 - 3x + 2\) type.
- Polynomial inequalities and the sign-chart (wavy-curve) method.
- Location of roots within intervals (e.g., both roots between \(-3\) and \(5\)).
- Common roots between two equations.
This topic connects directly to complex numbers (complex conjugate roots), sets, relations, and functions (domain/range via discriminant), and sequences and series (roots in AP or GP).
Exam Pattern & Weightage
The table below maps questions from the PYQ file to their exam year. NDA I and NDA II are treated separately where the paper is labelled.
| Year | No. | Dominant sub-topics |
|---|---|---|
| 2010 (I & II) | 5 | Roots of \(x^2 - 2x + 4 = 0\); condition for real roots; reducible equations |
| 2011 (I & II) | 12 | Reciprocal roots; double roots; inequalities; complex conjugate pair; sum/product |
| 2012 (I & II) | 6 | Roots scaled by 2; sum of squares of roots; common roots; symmetric roots |
| 2013 (I & II) | 10 | Reducible biquadratic; modulus equations; irrational coefficients; area via quadratic |
| 2014 (I & II) | 4 | Real roots guarantee; \(x^2 + px + 4 > 0\) for all real \(x\) |
| 2015 (I & II) | 6 | Student-mistake problem; modulus root count; equal sum of squares |
| 2016 (I & II) | 8 | Root shift by h; real-root conditions; location of roots; GP roots |
| 2017 (I & II) | 7 | Modulus sum; roots of \((q-r)x^2 + (r-p)x + (p-q) = 0\); cube roots of unity |
| 2018 (I & II) | 5 | Irrational-root detection; greatest-integer equation; ratio of roots |
| 2019 (I & II) | 12 | Modulus inequalities; root count; \(p\) and \(q\) from \(\{1,2,3,4\}\) real-root count; GP roots |
| 2020 I | 3 | \(\cot(\alpha+\beta)\); number of quadratic equations; inequality \((2x-3)(2x-9)\) |
| 2021 (I & II) | 8 | Reciprocal roots; roots differing by unity; equal-magnitude opposite-sign roots |
| 2022 (I & II) | 5 | \(\alpha^2 - \beta^2 = 16\) problem; sum = product question; \(p - q\) formula with \(c\) |
| 2023 (I) | 8 | Location in \((0,5)\); modulus equation; HM/GM of roots; equation \((1-x)^4+(5-x)^4=82\) |
NDA 2019 alone had 12 questions from this topic across both sittings — the highest single-year count in the dataset. Treat inequalities (modulus and polynomial) as equal priority to equation-solving; they now appear in almost every paper.
Core Concepts
The Standard Quadratic and its Discriminant
Every quadratic is \(ax^2 + bx + c = 0\), \(a \ne 0\). The nature of its roots is completely determined by \(D = b^2 - 4ac\):
\(D > 0\) → two distinct real roots
\(D = 0\) → two equal real roots (each \(= -b/2a\))
\(D < 0\) → two complex conjugate roots (no real roots)
When the paper says "roots are real", set \(D \ge 0\). When it says "roots are equal", set \(D = 0\). NDA regularly phrases these as "under what condition" questions — all you need is this one tool.
\(D\) is positive but not a perfect square → roots are irrational conjugates \(p \pm \sqrt{q}\)
\(D < 0\) → roots are complex conjugates \(p \pm iq\)
Complex roots always come in conjugate pairs when the coefficients are real. If a problem states one root is \(2 + 5i\), the other root is automatically \(2 - 5i\) — you don't need to solve anything. NDA 2011 and 2017 both tested this with cube-roots-of-unity setups.
For example, from the 2013 PYQ: the equation \(x^2 + bx + 4 = 0\) will have real roots if \(b \le -4\) or \(b \ge 4\), because \(D = b^2 - 16 \ge 0\) gives \(|b| \ge 4\).
Vieta's Formulas
If \(\alpha\) and \(\beta\) are roots of \(ax^2 + bx + c = 0\):
$$|\alpha - \beta| = \frac{\sqrt{D}}{|a|}$$
Most "value of expression" questions reduce to manipulating these two quantities. Key derived results you need:
$$\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$$
$$\alpha^2 - \beta^2 = (\alpha+\beta)(\alpha-\beta)$$
$$\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = -\dfrac{b}{c}$$
$$\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}$$
Don't confuse "sum of squares of roots" with "square of sum of roots". The correct expansion is \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\), not \((\alpha + \beta)^2\). NDA examiners regularly include the wrong expansion as a distractor option.
From the 2013 PYQ: if \(\alpha\) and \(\beta\) are roots of \(x^2 + bx + c = 0\), then \(\alpha^{-1} + \beta^{-1} = -b/c\).
From the 2019 PYQ: if \(p\) and \(q\) are roots of \(x^2 - 30x + 221 = 0\), then \(p + q = 30\), \(pq = 221\). To find \(p^3 + q^3 = (p+q)^3 - 3pq(p+q) = 27000 - 3(221)(30) = 27000 - 19890 = 7110\).
Forming a Quadratic from its Roots
i.e., $$x^2 - (\text{Sum})x + (\text{Product}) = 0$$
From NDA 2013: if sum \(= 3\) and product \(= 2\), the equation is \(x^2 - 3x + 2 = 0\).
When roots are \(m+n\) and \(m-n\) (NDA 2012): sum \(= 2m\), product \(= m^2 - n^2\), so the equation is \(x^2 - 2mx + (m^2 - n^2) = 0\).
Special Root Conditions
NDA tests these repeatedly — memorise the conditions, not just the formulas:
(Product of roots \(= c/a = 1\))
From NDA 2021: "one root of \(5x^2 + 26x + k = 0\) is reciprocal of the other" → \(k = a = 5\).
Equations Reducible to Quadratic
When you see \(x^4 - 26x^2 + 25 = 0\), substitute \(t = x^2\) to get \(t^2 - 26t + 25 = 0\), factor as \((t-1)(t-25) = 0\), giving \(t = 1\) or \(t = 25\), so \(x = \pm 1, \pm 5\). The solution set is \(\{-5, -1, 1, 5\}\) (NDA 2011).
Exponential form: \(4^x - 6 \cdot 2^x + 8 = 0\). Let \(t = 2^x \implies t^2 - 6t + 8 = 0 \implies (t-2)(t-4) = 0 \implies 2^x = 2\) or \(2^x = 4 \implies x = 1\) or \(x = 2\) (NDA 2013).
Modulus Equations and Inequalities
The modulus (absolute value) of a quadratic generates the most variety in recent papers. Key method: split into cases \(x \ge\) threshold and \(x <\) threshold, solve each, verify solutions.
\(f(x) > 0\) when \(x < a\) or \(x > b\)
\(f(x) < 0\) when \(a < x < b\)
From NDA 2019: if \(|x^2 - 3x + 2| > x^2 - 3x + 2\), then the expression inside is strictly negative, so \(x^2 - 3x + 2 < 0\), giving \((x-1)(x-2) < 0\), i.e., \(1 < x < 2\).
From NDA 2020: if \(1.5 < x < 4.5\), then \(2x - 3 > 0\) and \(2x - 9 < 0\), so \((2x-3)(2x-9) < 0\).
Location of Roots
For roots to lie in an interval \((p, q)\), you need: \(D \ge 0\), \(f(p)\) and \(f(q)\) have the same sign as \(a\), and \(p < -b/(2a) < q\). From NDA 2016: both roots of \(x^2 - 2kx + k^2 - 4 = 0\) lie between \(-3\) and \(5\), which forces \(-2 < k < 2\).
Quadratic Inequality: Always Positive / Always Negative
From NDA 2016: \(x^2 + px + 4 > 0\) for all real \(x\) requires \(D = p^2 - 16 < 0\), i.e., \(|p| < 4\).
When you multiply or divide an inequality by a negative quantity, the inequality sign flips. Before applying the wavy-curve method, always make the coefficient of \(x^2\) positive — if it isn't, multiply through by \(-1\) and reverse the sign. Forgetting this single step is the most common source of full-mark losses in inequality questions.
Parabola Vertex — Maximum and Minimum Values
The graph of $$y = ax^2 + bx + c$$ is a parabola. If $$a > 0$$ it opens upward (has a minimum); if $$a < 0$$ it opens downward (has a maximum). The turning point is the vertex.
Extreme value of expression $$= -\dfrac{D}{4a}$$ (minimum if $$a > 0$$, maximum if $$a < 0$$)
Worked Examples
Example 1 — Sum of Cubes of Roots (NDA 2019)
If \(p\) and \(q\) are roots of \(x^2 - 30x + 221 = 0\), what is \(p^3 + q^3\)?
- By Vieta: \(p + q = 30\), \(pq = 221\).
- Use identity: $$p^3 + q^3 = (p + q)^3 - 3pq(p + q)$$
- \(= (30)^3 - 3(221)(30) = 27000 - 19890\).
- \(= 7110\). Answer: (b) \(7110\).
Example 2 — Reciprocal of Other Root (NDA 2021)
One root of \(5x^2 + 26x + k = 0\) is the reciprocal of the other. Find \(k\).
- Let roots be \(r\) and \(1/r\). Product of roots \(= (r)(1/r) = 1\).
- By Vieta: product \(= k/5\). So \(k/5 = 1 \implies k = 5\).
- Answer: \(k = 5\).
Example 3 — Modulus Inequality (NDA 2019)
If \(|x^2 - 3x + 2| > x^2 - 3x + 2\), which values of \(x\) satisfy this?
- \(|A| > A\) holds exactly when \(A < 0\).
- So \(x^2 - 3x + 2 < 0 \implies (x - 1)(x - 2) < 0\).
- Sign chart: expression is negative for \(1 < x < 2\).
- Answer: \(1 < x < 2\).
Example 4 — Reducible Equation (NDA 2011)
What is the solution set of \(x^4 - 26x^2 + 25 = 0\)?
- Let \(t = x^2\). Equation becomes \(t^2 - 26t + 25 = 0\).
- Factor: \((t - 1)(t - 25) = 0 \implies t = 1\) or \(t = 25\).
- \(t = 1 \implies x = \pm 1\); \(t = 25 \implies x = \pm 5\).
- Solution set: \(\{-5, -1, 1, 5\}\). Answer: (a).
Example 5 — Roots of Opposite Sign (NDA 2011)
One root of \(ax^2 + bx + c = 0\) is positive and the other is negative. What condition ensures this?
- Product of roots \(= c/a\). For one root positive and one negative, product must be negative.
- So \(c/a < 0\), meaning \(a\) and \(c\) must have opposite signs.
- This matches option (d): \(a < 0, c > 0\) (or equivalently \(a > 0, c < 0\)).
- No restriction on \(b\) is needed beyond the opposite-sign condition on \(a\) and \(c\).
The "student-mistake" problem (NDA 2015) is a classic trap. One student misreads the constant term and gets roots \(8\) and \(2\) (product \(= 16\), correct). Another misreads the \(x\)-coefficient and gets roots \(-9\) and \(-1\) (sum \(= -10\), correct). Correct equation: \(x^2 - 10x + 16 = 0\). Always extract the "correct" coefficient from the student who did not make that mistake.
Example 6 — Nature of Roots from Discriminant
Find the values of $$k$$ for which $$kx^2 - 4x + k = 0$$ has equal roots.
- Equal roots ⇒ $$D = 0$$, i.e., $$b^2 - 4ac = 0$$.
- $$(-4)^2 - 4(k)(k) = 0 \Rightarrow 16 - 4k^2 = 0$$.
- $$k^2 = 4 \Rightarrow k = \pm 2$$.
- Answer: $$k = 2$$ or $$k = -2$$.
Example 7 — Symmetric Function without Solving
If $$\alpha$$ and $$\beta$$ are roots of $$2x^2 - 5x + 3 = 0$$, find $$\alpha^2 + \beta^2$$.
- By Vieta: $$\alpha + \beta = \dfrac{5}{2}$$ and $$\alpha\beta = \dfrac{3}{2}$$.
- Use identity: $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$.
- $$= \left(\dfrac{5}{2}\right)^2 - 2\left(\dfrac{3}{2}\right) = \dfrac{25}{4} - 3 = \dfrac{13}{4}$$.
- Answer: $$\dfrac{13}{4}$$. No need to find $$\alpha$$ or $$\beta$$ individually.
Example 8 — Quadratic Inequality by Wavy-Curve
Solve $$x^2 - 5x + 6 < 0$$.
- Coefficient of $$x^2$$ is already positive — no sign flip needed.
- Factorise: $$(x-2)(x-3) < 0$$. Critical points: 2 and 3.
- Sign chart: $$(-\infty, 2)$$ is $$+$$, $$(2, 3)$$ is $$-$$, $$(3, \infty)$$ is $$+$$.
- Inequality is $$< 0$$, so pick the negative region: $$x \in (2, 3)$$.
Exam Shortcuts (Pro-Tips)
Quadratic questions reward pattern recognition. The four shortcuts below collapse a full factoring or Vieta substitution into a single inspection. Each one has been tested in NDA papers — memorise the trigger condition, not just the formula.
Shortcut 1 — Cyclic Coefficient Trick ($$a + b + c = 0$$)
If the coefficients of $$ax^2 + bx + c = 0$$ sum to zero, then $$x = 1$$ is automatically a root (substitute and check). The other root is $$\dfrac{c}{a}$$ (from product of roots).
Example: For $$7x^2 - 10x + 3 = 0$$, check $$7 - 10 + 3 = 0$$. Roots are $$1$$ and $$\dfrac{3}{7}$$ instantly — no factoring required.
Shortcut 2 — Alternate-Sign Trick ($$a - b + c = 0$$)
If $$a - b + c = 0$$ (equivalently $$a + c = b$$), then $$x = -1$$ is a root. The other root is $$-\dfrac{c}{a}$$.
Shortcut 3 — Coefficient-Driven Root Patterns
Three "by-inspection" patterns NDA repeatedly tests:
$$b = 0$$ → roots are equal in magnitude, opposite sign: $$x = \pm\sqrt{-c/a}$$
$$a = c$$ → roots are reciprocals: $$\alpha$$ and $$\dfrac{1}{\alpha}$$
NDA 2021 tested the third pattern directly: "one root of $$5x^2 + 26x + k = 0$$ is the reciprocal of the other" → $$k = a = 5$$. Solved in 3 seconds without touching the quadratic formula.
Shortcut 4 — Sign Analysis for "Always Positive / Always Negative"
For $$ax^2 + bx + c$$ to maintain one sign across all real $$x$$, the parabola must not cross the x-axis. Combine the leading-coefficient sign with $$D < 0$$:
Always negative ⇔ $$a < 0$$ and $$D < 0$$
NDA 2014 / 2016: $$x^2 + px + 4 > 0$$ for all real $$x$$ → $$a = 1 > 0$$ ✓, need $$D = p^2 - 16 < 0$$, so $$|p| < 4$$, i.e., $$p \in (-4, 4)$$.
Shortcut 5 — Common-Root Test (Two Equations)
If the equations share both roots, their coefficients are proportional: $$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$$. For one common root, the cross-multiplication condition gives:
In practice, subtract the two equations to eliminate $$x^2$$ — the resulting linear equation gives the common root directly. Faster than memorising the determinant form.
Common Question Patterns
After analysing over 130 questions from the PYQ file, NDA quadratic questions cluster into six recognisable patterns:
Pattern 1 — Nature of Roots
You are given a parametric equation and asked for the range of the parameter that ensures real, equal, distinct, or complex roots. Always write \(D \ge 0\) (or \(D = 0\) or \(D < 0\)) and solve for the parameter. Seen in: 2010, 2011, 2013, 2014, 2016, 2019, 2021, 2023.
Pattern 2 — Symmetric Functions of Roots
You are given \(\alpha + \beta\) and \(\alpha\beta\), then asked for \(\alpha^2 + \beta^2\), \(\alpha^3 + \beta^3\), \(\alpha^{-1} + \beta^{-1}\), or products of linear expressions in roots. Express everything through the sum and product; never try to find the individual roots. Seen in: 2010, 2011, 2013, 2017, 2019.
Pattern 3 — Special Root Relationships
One root is double the other, roots differ by unity, roots are reciprocal, roots are equal, one root is \(1\), etc. Each has a dedicated condition formula — memorise them. Seen in: 2011, 2013, 2015, 2021.
Pattern 4 — Reducible Equations
Biquadratic (\(x^4\) form), exponential (\(4^x\) form), trigonometric (\(\tan^2 x\) form), or greatest-integer equations. The key move is always to substitute to convert the equation to standard quadratic form. Seen in: 2011, 2013, 2018.
Pattern 5 — Modulus Equations and Inequalities
\(|f(x)| = g(x)\) or \(|f(x)| > f(x)\) type. Split into cases: \(f(x) \ge 0\) and \(f(x) < 0\). For \(|f(x)| > f(x)\), the answer is always \(f(x) < 0\), solved by sign chart. Seen in: 2013, 2015, 2017, 2019.
Pattern 6 — Location of Roots in an Interval
Both roots between two values, or exactly one root in an interval. Use: \(D \ge 0\), sign of \(f\) at endpoints, vertex \(x\)-coordinate in interval. Seen in: 2016, 2023.
How NDA Tests This Topic
About 40% of questions in this topic are single-step: plug into Vieta and evaluate. Another 35% need a two-step manipulation (like \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)). Only the remaining 25% — location of roots, modulus inequalities, reducible equations — require three or more steps. Speed matters: if you cannot solve a single-step Vieta question in 30 seconds, your base recall needs work before you focus on the harder patterns.
Preparation Strategy
Step 1 — Lock in the Formulas (Days 1–2)
Write out the discriminant rule, Vieta's sum and product, and the five derived expressions (\(\alpha^2 + \beta^2\), \(\alpha^3 + \beta^3\), etc.) on a single flash card. Test yourself without looking until you can produce them in under 10 seconds each.
Step 2 — Drill Special Conditions (Days 3–4)
Reciprocal roots (\(a = c\)), double root (\(2b^2 = 9ac\)), roots differ by \(1\) (\(b^2 = a^2 + 4ac\)), opposite-sign roots (\(c/a < 0\)), always-positive quadratic (\(a > 0, D < 0\)). Solve one PYQ per condition per day — the 2011 and 2021 papers have the richest set of these questions.
Step 3 — Modulus and Inequalities (Days 5–6)
Practice the sign-chart method for polynomial inequalities. For modulus equations, drill the case-split technique until it is automatic. Refer to NDA 2019 questions on \(|x^2 - 3x + 2|\) and \((2x-3)(2x-9)\) as model problems.
Step 4 — Full PYQ Pass (Day 7 onwards)
Work through all 130+ questions chronologically. Mark any question where you took more than 2 minutes or made an error. Group errors by pattern and spend extra time on whichever pattern has the most red marks. Use mock tests to simulate NDA time pressure — 120 questions in 150 minutes leaves just 75 seconds per question.
Common Errors to Avoid
- Forgetting that "real and equal roots" means \(D = 0\), not \(D \ge 0\).
- Applying reciprocal-root condition as \(b = 0\) instead of \(a = c\).
- In modulus questions, not verifying solutions back in the original equation — extraneous roots appear often.
- In reducible equations, solving for the substituted variable but forgetting to solve back for \(x\) (missing \(\pm\) signs).
- Misreading "sum of squares of roots" as \((\alpha + \beta)^2\) instead of \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).
For broader algebraic context, revisit sets, relations, and functions for domain/range reasoning that overlaps with quadratic inequalities, and check binomial theorem and logarithms for questions where logarithmic discriminant conditions appear (as in the NDA 2011 question on \(x^2 - 4x - \log_3 N = 0\)).
Test Your Quadratic Skills Right Now
Free NDA mock tests with full-length papers, timer, and topic-wise analytics. See exactly how many marks this topic is worth for you.
Start Free Mock TestFrequently Asked Questions
How many questions from quadratic equations appear in one NDA paper?
Based on the PYQ data from 2010 to 2023, the count ranges from 3 to 12 questions per paper sitting, with an average of around 5–6. NDA 2019 had the highest single-year total. You can expect at least 4 questions in every sitting, making this one of the most consistent scoring areas in Paper I Maths.
What is the most important formula to know for NDA quadratic questions?
Vieta's formulas: sum of roots \(= -b/a\) and product of roots \(= c/a\). Roughly 70% of NDA questions in this topic either use these directly or use expressions derived from them (\(\alpha^2 + \beta^2\), \(\alpha^3 + \beta^3\), \(\alpha^{-1} + \beta^{-1}\)). If you know nothing else, know these and the discriminant condition.
What does it mean when NDA asks for the "condition" for real roots?
It means you must set the discriminant \(D = b^2 - 4ac \ge 0\) (for real roots, possibly equal) or \(D > 0\) (for distinct real roots). Solve the resulting inequality for the parameter (often labelled \(k, p, m\), etc.). The answer will be a range or a specific value.
How do I handle equations like x⁴ − 26x² + 25 = 0 in NDA?
Substitute \(t = x^2\) (or whatever the repeated variable is). The equation becomes a standard quadratic in \(t\). Solve for \(t\), then substitute back to find \(x\). Remember to take both positive and negative square roots when going back to \(x\) — NDA questions like this (2011) test whether you include all four roots in the solution set.
When should I use the sign-chart (wavy-curve) method?
Use it whenever you have a polynomial inequality like \((x - a)(x - b) > 0\) or a modulus expression. Mark the roots on a number line, then alternate signs starting from the rightmost region (positive for even-degree leading coefficient). It is faster than testing individual values, especially when three or more factors appear.
What is the condition for one root to be the reciprocal of the other?
The product of roots must equal \(1\). By Vieta, product \(= c/a\), so \(c/a = 1\) gives \(c = a\). This appeared in NDA 2011 and again in NDA 2021 (\(5x^2 + 26x + k = 0 \implies k = 5\)). Never confuse this with the condition \(b = 0\), which means the sum of roots is zero (roots are negatives of each other, not reciprocals).
How is this topic related to complex numbers in NDA?
When \(D < 0\), the roots are complex conjugates \(a \pm bi\). NDA tests this through cube roots of unity (\(\omega, \omega^2\)) which satisfy \(x^2 + x + 1 = 0\), and through problems asking you to recognise that if one root is \(2 + 5i\) then the other is \(2 - 5i\) (NDA 2011). Study the complex numbers topic alongside this one for complete coverage.