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.

Concept 1 of 6

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

Intuition

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=ωnmod3\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}.
  1. Factor out the lowest power: x16(1+x+x2)x^{16}(1 + x + x^2).
  2. Since xx is a root, 1+x+x2=01 + x + x^2 = 0.
  3. So the whole product is x160x^{16} \cdot 0.
Answer:00.

From the bank · past-year question

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

[Q1 · Sep · 2021]

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 ωnmod3\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

Intuition

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 (qr)x2+(rp)x+(pq)=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 pqqr\dfrac{p-q}{q-r} (from 1β=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

(qr)x2+(rp)x+(pq)=0  x=1,  x=pqqr(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 (bc)x2+(ca)x+(ab)=0(b-c)x^2 + (c-a)x + (a-b) = 0.
  1. Coefficients sum to zero: (bc)+(ca)+(ab)=0(b-c)+(c-a)+(a-b) = 0, so x=1x = 1 is a root.
  2. Product of roots =abbc= \dfrac{a-b}{b-c}, and one root is 11.
Answer:x=1x = 1 and x=abbcx = \dfrac{a-b}{b-c}.

From the bank · past-year question

Example 2Quadratic EquationsMODERATE
The roots of the equation (qr)x2+(rp)x+(pq)=0(q-r)x^2 + (r-p)x + (p-q) = 0 are

[Q5 · Sep · 2017]

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

Intuition

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 xa2+xa2=0|x-a|^2 + |x-a| - 2 = 0, set t=xa0t = |x-a| \ge 0, solve t2+t2=0t^2 + t - 2 = 0, keep only t0t \ge 0, then undo t=xat = |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+kx+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

xa2+xa2=0,  t=xa0  t2+t2=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 x32+x32=0|x-3|^2 + |x-3| - 2 = 0.
  1. Let t=x30t = |x-3| \ge 0: t2+t2=(t+2)(t1)=0t^2 + t - 2 = (t+2)(t-1) = 0, so t=1t = 1 (reject t=2t = -2).
  2. x3=1x=4|x-3| = 1 \Rightarrow x = 4 or x=2x = 2.
  3. Sum =4+2= 4 + 2.
Answer:66.

From the bank · past-year question

Example 3Quadratic EquationsMODERATE
What are the roots of the equation x2x6=x+2|x^2 - x - 6| = x + 2?

[Q8 · Apr · 2019]

A negative value of the modulus variable is impossible

After t=xat = |x-a|, discard any negative tt before solving back. And for f=g|f| = g, a candidate root is valid only if g0g \ge 0 there — skip this and you import roots the equation never had.

Concept 4 of 6

Parametric Quadratics — Factor, Don't Force

Intuition

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)xc](x1)=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 =cb24a=k2k24=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)=cb24a=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.
  1. Try to factor with x=1x = 1: coefficients sum to (a+b)(a+b+c)+c=0(a+b) - (a+b+c) + c = 0, so x=1x = 1 is a root.
  2. Factor: [(a+b)xc](x1)=0[(a+b)x - c](x-1) = 0.
Answer:x=1x = 1 and x=ca+bx = \dfrac{c}{a+b}.

From the bank · past-year question

Example 4Quadratic EquationsMODERATE
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:

[Q37 · Sep · 2023]

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

Intuition

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 logb()\log_b(\,\cdot\,), substitute t=logb(argument)t = \log_b(\text{argument}), solve the quadratic in tt, then convert back with t=logbu    u=btt = \log_b u \iff u = b^{\,t}. Useful log facts: logbu=1logub\log_b u = \dfrac{1}{\log_u b} (reciprocal of base and argument) and logb2u=12logbu\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=logbu  quadratic in t,u=btt = \log_b u \ \Rightarrow\ \text{quadratic in } t,\quad u = b^{\,t}

Worked example

Solve (log10x)23log10x+2=0(\log_{10} x)^2 - 3\log_{10} x + 2 = 0.
  1. Let t=log10xt = \log_{10} x: t23t+2=(t1)(t2)=0t^2 - 3t + 2 = (t-1)(t-2) = 0, so t=1t = 1 or t=2t = 2.
  2. Undo the log: log10x=1x=10\log_{10} x = 1 \Rightarrow x = 10; log10x=2x=100\log_{10} x = 2 \Rightarrow x = 100.
Answer:x=10x = 10 or x=100x = 100.

From the bank · past-year question

Example 5Quadratic EquationsHARD
For the following two (02) items: Let α\alpha and β\beta be the roots of the quadratic equation x2+(log0.5(a2))x+(log0.5(a2))4=0x^2+(\log_{0.5}(a^2))x+(\log_{0.5}(a^2))^4=0 where a21a^2\neq 1 and log0.5(a2)>0\log_{0.5}(a^2)>0. Further, β2=α(loga2(0.5))\beta^2=\alpha(\log_{a^2}(0.5)).
What is β\beta equal to?

[Q21 · Sep · 2025]

Solve for the log first, the variable second

The quadratic is in t=logxt = \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

Intuition

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:

  • x2axbx+(abc)=0x^2 - ax - bx + (ab - c) = 0 is x2(a+b)x+(abc)=0x^2 - (a+b)x + (ab-c) = 0, so α+β=a+b\alpha+\beta = a+b and αβ=abc\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 x2Sx+P=0x^2 - Sx + P = 0. E.g. if α3,β3\alpha^3, \beta^3 are the new roots, S=(α+β)33αβ(α+β)S = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) and P=(αβ)3P = (\alpha\beta)^3.

Expand, then Vieta

x2axbx+(abc)=0  α+β=a+b,  αβ=abcx^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.
  1. From x2+px+q=0x^2+px+q=0: α+β=p\alpha+\beta = -p, αβ=q\alpha\beta = q.
  2. New sum =α3+β3=(α+β)33αβ(α+β)=(p)33q(p)= \alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = (-p)^3 - 3q(-p).
Answer:p3+3pq-p^3 + 3pq.

From the bank · past-year question

Example 6Quadratic EquationsHARD
Let α\alpha and β\beta be the roots of the equation x2axbx+abc=0x^2-ax-bx+ab-c=0. What is the quadratic equation whose roots are aa and bb?

[Q18 · Apr · 2022]

Expand the constructed form before reading coefficients

x2axbx+abcx^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

    (qr)x2+(rp)x+(pq)=0  x=1,  x=pqqr(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

    xa2+xa2=0,  t=xa0  t2+t2=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)=cb24a=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=logbu  quadratic in t,u=btt = \log_b u \ \Rightarrow\ \text{quadratic in } t,\quad u = b^{\,t}
  • Quadratics Built From Their Roots

    Expand, then Vieta

    x2axbx+(abc)=0  α+β=a+b,  αβ=abcx^2 - ax - bx + (ab - c) = 0 \ \Rightarrow\ \alpha+\beta = a+b,\ \ \alpha\beta = ab - c

Watch out for (6)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Quadratic EquationsMODERATE
If kk is one of the roots of the equation x(x+1)+1=0x(x+1)+1=0, then what is its other root?

[Q113 · Apr · 2021]

Example 2Quadratic EquationsHARD
The roots of the quadratic equation (a2b2c2)x2+(b2c2a2)x+(c2a2b2)=0(a^2b^2-c^2)x^2+(b^2c^2-a^2)x+(c^2a^2-b^2)=0 are equal, a2b2c2a^2\neq b^2\neq c^2.
Which one of the following is a root of the equation?

[Q48 · Sep · 2024]

Example 3Quadratic EquationsEASY
The number of real roots for the equation x2+9x+20=0x^2+9|x|+20=0 is

[Q78 · Apr · 2019]

Example 4Quadratic EquationsHARD
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=c/2 (c0)k=c/2\ (c\neq0), then the roots of the equation are:

[Q36 · Sep · 2023]

Example 5Quadratic EquationsHARD
For the following two (02) items: Let α\alpha and β\beta be the roots of the quadratic equation x2+(log0.5(a2))x+(log0.5(a2))4=0x^2+(\log_{0.5}(a^2))x+(\log_{0.5}(a^2))^4=0 where a21a^2\neq 1 and log0.5(a2)>0\log_{0.5}(a^2)>0. Further, β2=α(loga2(0.5))\beta^2=\alpha(\log_{a^2}(0.5)).
What is the relation between α\alpha and β\beta?

[Q22 · Sep · 2025]

Drill every past-year question on this subtopic

16 questions from the bank — paginated, with cart and Word-export support.