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NDA Maths · Quadratic Equations

Special Quadratics — Parametric, Logarithmic & Constructed

A family of disguised quadratics: cube-roots-of-unity hooks, modulus and logarithmic equations that reduce to a quadratic, and equations built from a parameter or from their own roots.

Why this matters

16 PYQs, 7 of them HARD — this subtopic is where the chapter's hardest, most disguised questions live. The recurring star is x²+x+1=0, the gateway to the cube roots of unity ω (which also power the ω+Vieta compound). The rest reward one reflex: spot the substitution — split the modulus, name the log, factor the parameter — that turns the disguise back into an ordinary quadratic.

Test yourself — a quick 17-question Quadratic Equations recall quiz, instant score.

Concept 1 of 6: Cube Roots of Unity — the x² + x + 1 Hook

Whenever you see x² + x + 1 = 0 (or x(x+1)+1 = 0), its roots are the complex cube roots of unity ω and ω². They satisfy two tiny facts — ω³ = 1 and 1 + ω + ω² = 0 — that collapse enormous powers to almost nothing.

Definition

The roots of x2+x+1=0x^2 + x + 1 = 0 are the non-real cube roots of unity ω\omega and ω2\omega^2, with:

  • ω3=1\omega^3 = 1 (so powers cycle with period 3: ωn=ω n mod 3\omega^n = \omega^{\,n \bmod 3}),
  • 1+ω+ω2=01 + \omega + \omega^2 = 0,
  • ω2=ωˉ=1ω\omega^2 = \bar{\omega} = \dfrac{1}{\omega} (the two roots are each other's square and reciprocal).

So if xx is a root of x2+x+1=0x^2+x+1=0: any x3k=1x^{3k} = 1, and a sum like xn+xn+1+xn+2=xn(1+x+x2)=0x^n + x^{n+1} + x^{n+2} = x^n(1 + x + x^2) = 0.

The two defining facts

ω3=1,1+ω+ω2=0\omega^3 = 1, \qquad 1 + \omega + \omega^2 = 0

Worked example

If x2+x+1=0x^2 + x + 1 = 0, find x16+x17+x18x^{16} + x^{17} + x^{18}.

The same idea in a real exam question:

NDA · Sep 2021 · Q1Moderate

Example 1 · Quadratic Equations · Special Quadratics — Parametric, Logarithmic, Constructed

If x2+x+1=0x^{2} + x + 1 = 0, then what is the value of x199+x200+x201x^{199} + x^{200} + x^{201}?

Reduce the exponent mod 3 before anything else

ω200\omega^{200} is not a monster — 200=3(66)+2200 = 3(66) + 2, so ω200=ω2\omega^{200} = \omega^2. Always replace ωn\omega^n by ωn mod 3\omega^{n \bmod 3} first, and look for the 1+x+x21+x+x^2 factor to send a sum to zero.

Concept 2 of 6: Constructed Symmetric-Coefficient Equations

When the coefficients are built symmetrically from letters — (q−r), (r−p), (p−q) — they are engineered so the coefficients sum to zero, handing you x = 1 as a free root. The same constructions, set to have equal roots, fold back to the AP/GP/HP conditions from Subtopic 1.

Definition

Two moves for symbol-coefficient quadratics:

  • Sum-to-zero ⇒ unit root. For (q−r)x2+(r−p)x+(p−q)=0(q-r)x^2 + (r-p)x + (p-q) = 0, the coefficients add to 00, so x=1x = 1 is a root and the other root is the product p−qq−r\dfrac{p-q}{q-r} (from 1⋅β=c/a1 \cdot \beta = c/a).
  • Equal roots ⇒ a progression. A constructed equation whose roots are equal yields D=0D = 0, which simplifies to an AP/GP/HP relation among the letters — often the HP form 2b2=1a2+1c2\frac{2}{b^2} = \frac{1}{a^2} + \frac{1}{c^2} for the squared-coefficient family. The repeated root is x=−B2Ax = -\dfrac{B}{2A} (not −B/A-B/A).

Symmetric construction ⇒ unit root

(q−r)x2+(r−p)x+(p−q)=0 ⇒ x=1,  x=p−qq−r(q-r)x^2 + (r-p)x + (p-q) = 0 \ \Rightarrow\ x = 1,\ \ x = \tfrac{p-q}{q-r}

Worked example

Find both roots of (b−c)x2+(c−a)x+(a−b)=0(b-c)x^2 + (c-a)x + (a-b) = 0.

The same idea in a real exam question:

NDA · Sep 2017 · Q5Moderate

Example 2 · Quadratic Equations · Special Quadratics — Parametric, Logarithmic, Constructed

The roots of the equation (q−r)x2+(r−p)x+(p−q)=0(q-r)x^2 + (r-p)x + (p-q) = 0 are

Repeated root is −B/2A, not −B/A

For equal roots the single root equals the vertex x=−B2Ax = -\frac{B}{2A} (half of the sum of roots, since both roots coincide). Using −B/A-B/A (the full sum) doubles it — the offered wrong answer.

Concept 3 of 6: Modulus Equations Reducing to Quadratics

An absolute value hides a quadratic. Either substitute the modulus as a single non-negative variable, or split into the two sign cases — but always finish by checking each candidate against the domain the case came from.

Definition

Two strategies for an equation containing ∣ ⋅ ∣|\,\cdot\,|:

  • Substitute the modulus: for ∣x−a∣2+∣x−a∣−2=0|x-a|^2 + |x-a| - 2 = 0, set t=∣x−a∣≥0t = |x-a| \ge 0, solve t2+t−2=0t^2 + t - 2 = 0, keep only t≥0t \ge 0, then undo t=∣x−a∣t = |x-a|.
  • Split by sign: for ∣f(x)∣=g(x)|f(x)| = g(x), first require g(x)≥0g(x) \ge 0, then solve f(x)=g(x)f(x) = g(x) and f(x)=−g(x)f(x) = -g(x), keeping only roots that satisfy the sign assumption.

Note x2+k∣x∣+m=0x^2 + k|x| + m = 0 with k,m>0k, m > 0 has no real root — a sum of non-negative terms can't vanish.

Substitute t = |·| ≥ 0

∣x−a∣2+∣x−a∣−2=0,  t=∣x−a∣≥0 ⇒ t2+t−2=0|x-a|^2 + |x-a| - 2 = 0,\ \ t = |x-a| \ge 0 \ \Rightarrow\ t^2 + t - 2 = 0

Worked example

Find the sum of the real roots of ∣x−3∣2+∣x−3∣−2=0|x-3|^2 + |x-3| - 2 = 0.

The same idea in a real exam question:

NDA · Apr 2019 · Q8Moderate

Example 3 · Quadratic Equations · Special Quadratics — Parametric, Logarithmic, Constructed

What are the roots of the equation ∣x2−x−6∣=x+2|x^2 - x - 6| = x + 2?

A negative value of the modulus variable is impossible

After t=∣x−a∣t = |x-a|, discard any negative tt before solving back. And for ∣f∣=g|f| = g, a candidate root is valid only if g≥0g \ge 0 there — skip this and you import roots the equation never had.

Concept 4 of 6: Parametric Quadratics — Factor, Don't Force

When a quadratic carries a parameter and you're given a specific value of it, the cleaner path is almost always to factor the resulting expression — it usually splits neatly — rather than grinding the quadratic formula through messy symbols.

Definition

For a parametric quadratic at a given parameter value:

  • Substitute the value, then factor. E.g. with k=ck = c, (a+b)x2−(a+b+c)x+c=0(a+b)x^2 - (a+b+c)x + c = 0 factors as [(a+b)x−c](x−1)=0[(a+b)x - c](x - 1) = 0, giving x=1x = 1 and x=ca+bx = \dfrac{c}{a+b} directly — no formula needed.
  • Minimum of a parametric quadratic: x2+kx+k2x^2 + kx + k^2 has minimum value at the vertex =c−b24a=k2−k24=3k24= c - \dfrac{b^2}{4a} = k^2 - \dfrac{k^2}{4} = \dfrac{3k^2}{4}.
  • For a general parameter, the nature of the roots still comes from the discriminant in terms of that parameter.

Vertex (minimum) value, a > 0

min⁡(ax2+bx+c)=c−b24a=−D4a\min(ax^2+bx+c) = c - \dfrac{b^2}{4a} = -\dfrac{D}{4a}

Worked example

With k=ck = c, solve (a+b)x2−(a+b+c)x+c=0(a+b)x^2 - (a+b+c)x + c = 0.

The same idea in a real exam question:

NDA · Sep 2023 · Q37Moderate

Example 4 · Quadratic Equations · Special Quadratics — Parametric, Logarithmic, Constructed

A quadratic equation is given by (a+b)x2−(a+b+c)x+k=0(a+b)x^2-(a+b+c)x+k=0, where a,b,ca,b,c are real.
If k=ck=c, then the roots of the equation are:

Look for the unit-root before the formula

Parametric quadratics in NDA are nearly always built to factor (often with x=1x=1 a root). Reaching for −b±D2a\frac{-b\pm\sqrt{D}}{2a} with symbolic a,b,ca,b,c is slow and error-prone — test the coefficient sum first.

Concept 5 of 6: Logarithmic Equations That Are Quadratics

An equation that is quadratic in a logarithm is just an ordinary quadratic in disguise — name the log as a single variable, solve the quadratic, then undo the log. The only extra care is the domain (the argument of a log must be positive).

Definition

If an equation is quadratic in log⁡b( ⋅ )\log_b(\,\cdot\,), substitute t=log⁡b(argument)t = \log_b(\text{argument}), solve the quadratic in tt, then convert back with t=log⁡bu  ⟺  u=b tt = \log_b u \iff u = b^{\,t}. Useful log facts: log⁡bu=1log⁡ub\log_b u = \dfrac{1}{\log_u b} (reciprocal of base and argument) and log⁡b2u=12log⁡bu\log_{b^2} u = \tfrac{1}{2}\log_b u. When the coefficients of the quadratic are themselves logs, the same substitution turns Vieta's relations into relations among the logs.

Name the log, solve, invert

t=log⁡bu ⇒ quadratic in t,u=b tt = \log_b u \ \Rightarrow\ \text{quadratic in } t,\quad u = b^{\,t}

Worked example

Solve (log⁡10x)2−3log⁡10x+2=0(\log_{10} x)^2 - 3\log_{10} x + 2 = 0.

The same idea in a real exam question:

NDA · Sep 2025 · Q21Hard

Example 5 · Quadratic Equations · Special Quadratics — Parametric, Logarithmic, Constructed

For the following two (02) items: Let α\alpha and β\beta be the roots of the quadratic equation x2+(log⁡0.5(a2))x+(log⁡0.5(a2))4=0x^2+(\log_{0.5}(a^2))x+(\log_{0.5}(a^2))^4=0 where a2≠1a^2\neq 1 and log⁡0.5(a2)>0\log_{0.5}(a^2)>0. Further, β2=α(log⁡a2(0.5))\beta^2=\alpha(\log_{a^2}(0.5)).
What is β\beta equal to?

Solve for the log first, the variable second

The quadratic is in t=log⁡xt = \log x, not in xx. Its roots are values of the LOG; you still have to exponentiate to recover xx. And every recovered xx must keep the original log arguments positive.

Concept 6 of 6: Quadratics Built From Their Roots

An equation handed to you in a factored or constructed form already shows its sum and product once you expand it — read those off, then build whatever new equation the question wants by computing the new sum and product.

Definition

Expand a constructed quadratic to standard form and read off Vieta's relations:

  • x2−ax−bx+(ab−c)=0x^2 - ax - bx + (ab - c) = 0 is x2−(a+b)x+(ab−c)=0x^2 - (a+b)x + (ab-c) = 0, so α+β=a+b\alpha+\beta = a+b and αβ=ab−c\alpha\beta = ab - c.
  • To build the equation whose roots are a transform of α,β\alpha, \beta (their cubes, shifts, reciprocals), compute the new sum SS and product PP from α+β\alpha+\beta and αβ\alpha\beta, then write x2−Sx+P=0x^2 - Sx + P = 0. E.g. if α3,β3\alpha^3, \beta^3 are the new roots, S=(α+β)3−3αβ(α+β)S = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) and P=(αβ)3P = (\alpha\beta)^3.

Expand, then Vieta

x2−ax−bx+(ab−c)=0 ⇒ α+β=a+b,  αβ=ab−cx^2 - ax - bx + (ab - c) = 0 \ \Rightarrow\ \alpha+\beta = a+b,\ \ \alpha\beta = ab - c

Worked example

α,β\alpha, \beta are the roots of x2+px+q=0x^2 + px + q = 0. Find the sum of the roots of the equation whose roots are α3,β3\alpha^3, \beta^3.

The same idea in a real exam question:

NDA · Apr 2022 · Q18Hard

Example 6 · Quadratic Equations · Special Quadratics — Parametric, Logarithmic, Constructed

Let α\alpha and β\beta be the roots of the equation x2−ax−bx+ab−c=0x^2-ax-bx+ab-c=0. What is the quadratic equation whose roots are aa and bb?

Expand the constructed form before reading coefficients

x2−ax−bx+ab−cx^2 - ax - bx + ab - c looks like it has linear coefficient −a-a or −b-b; it doesn't until you combine them into −(a+b)-(a+b). Always collect like terms to standard form before applying Vieta.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

  • Cube Roots of Unity — the x² + x + 1 Hook

    The two defining facts

    ω3=1,1+ω+ω2=0\omega^3 = 1, \qquad 1 + \omega + \omega^2 = 0
  • Constructed Symmetric-Coefficient Equations

    Symmetric construction ⇒ unit root

    (q−r)x2+(r−p)x+(p−q)=0 ⇒ x=1,  x=p−qq−r(q-r)x^2 + (r-p)x + (p-q) = 0 \ \Rightarrow\ x = 1,\ \ x = \tfrac{p-q}{q-r}
  • Modulus Equations Reducing to Quadratics

    Substitute t = |·| ≥ 0

    ∣x−a∣2+∣x−a∣−2=0,  t=∣x−a∣≥0 ⇒ t2+t−2=0|x-a|^2 + |x-a| - 2 = 0,\ \ t = |x-a| \ge 0 \ \Rightarrow\ t^2 + t - 2 = 0
  • Parametric Quadratics — Factor, Don't Force

    Vertex (minimum) value, a > 0

    min⁡(ax2+bx+c)=c−b24a=−D4a\min(ax^2+bx+c) = c - \dfrac{b^2}{4a} = -\dfrac{D}{4a}
  • Logarithmic Equations That Are Quadratics

    Name the log, solve, invert

    t=log⁡bu ⇒ quadratic in t,u=b tt = \log_b u \ \Rightarrow\ \text{quadratic in } t,\quad u = b^{\,t}
  • Quadratics Built From Their Roots

    Expand, then Vieta

    x2−ax−bx+(ab−c)=0 ⇒ α+β=a+b,  αβ=ab−cx^2 - ax - bx + (ab - c) = 0 \ \Rightarrow\ \alpha+\beta = a+b,\ \ \alpha\beta = ab - c

Watch out for (6)

Test yourself on Quadratic Equations

15 past NDA questions from this chapter, timed at 19 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.