NDA Maths · Quadratic Equations

Vieta's Relations & Root-Coefficient Identities

Vieta's relations tie the sum and product of the roots directly to the coefficients — so almost any question about the roots can be answered from a, b, c without ever finding the roots themselves.

Why this matters

This is the chapter's centre of gravity — 26 PYQs, 11 of them HARD. The pattern is relentless: a question asks for a symmetric expression in the roots (their squares, cubes, reciprocals), or to build a new equation from transformed roots, and the whole thing collapses to sum and product. It also powers the ω+Vieta compound. Internalise α+β = −b/a and αβ = c/a together with the symmetric-function identities and most of these are one-liners.

Concept 1 of 7

Vieta's Relations — Sum and Product of Roots

Intuition

Expanding a(x−α)(x−β) and matching it to ax²+bx+c forces two identities: the sum of the roots is −b/a and the product is c/a. They are the workhorse of the whole chapter.

Definition

If α,β\alpha, \beta are the roots of ax2+bx+c=0ax^2 + bx + c = 0, then

α+β=ba,αβ=ca.\alpha + \beta = -\dfrac{b}{a}, \qquad \alpha\beta = \dfrac{c}{a}.
Consequences worth memorising:

  • Reciprocal roots (one root =1/= 1/other)     αβ=1    c=a\iff \alpha\beta = 1 \iff c = a.
  • Roots equal in magnitude, opposite sign     α+β=0    b=0\iff \alpha+\beta = 0 \iff b = 0.
  • **Sum of roots == product of roots**     b=c\iff -b = c.
  • Sign reading: if a,b,ca,b,c are all positive then α+β<0\alpha+\beta<0 and αβ>0\alpha\beta>0, so both roots are negative.

Vieta's relations

α+β=ba,αβ=ca\alpha + \beta = -\dfrac{b}{a}, \qquad \alpha\beta = \dfrac{c}{a}

Worked example

One root of 3x210x+k=03x^2 - 10x + k = 0 is the reciprocal of the other. Find kk.
  1. Reciprocal roots means the product of roots is 11.
  2. Product =ca=k3= \dfrac{c}{a} = \dfrac{k}{3}, so k3=1\dfrac{k}{3} = 1.
Answer:k=3k = 3.
Practice this concept2 quick reps

Practice — Level 1 (2 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Roots of x27x+12=0x^2 - 7x + 12 = 0: sum and product?
  2. 2.
    For what bb are the roots of x2+bx+9=0x^2 + bx + 9 = 0 equal in magnitude, opposite in sign?

From the bank · past-year question

Example 1Quadratic EquationsEASY
If pp and qq are the non-zero roots of the equation x2+px+q=0x^{2}+px+q=0, then how many possible values can qq have?

[Q8 · Sep · 2021]

Concept 2 of 7

Symmetric Functions & Forming New Equations

Intuition

Any expression in the roots that is unchanged when you swap α and β — their squares, cubes, difference, reciprocals — can be rewritten using only the sum s and product p. Compute the new sum and new product, and you can even build the quadratic whose roots are the transformed ones.

Definition

Let s=α+βs = \alpha+\beta, p=αβp = \alpha\beta. The standard identities:

  • α2+β2=s22p\alpha^2 + \beta^2 = s^2 - 2p
  • (αβ)2=s24p(\alpha-\beta)^2 = s^2 - 4p
  • α3+β3=s33ps=s(s23p)\alpha^3 + \beta^3 = s^3 - 3ps = s(s^2 - 3p)
  • 1α+1β=sp\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{s}{p}

Forming a new equation: if the new roots have sum SS and product PP, the quadratic is x2Sx+P=0x^2 - Sx + P = 0. Compute SS and PP as symmetric functions of α,β\alpha,\beta.

Build the equation from new sum & product

x2Sx+P=0,S=(new sum), P=(new product)x^2 - S x + P = 0, \quad S = \text{(new sum)},\ P = \text{(new product)}

Worked example

If p,qp, q are the roots of x25x+2=0x^2 - 5x + 2 = 0, find p3+q3p^3 + q^3.
  1. Sum s=5s = 5, product pq=2pq = 2.
  2. p3+q3=s(s23p)=5(256)=5×19p^3 + q^3 = s(s^2 - 3p) = 5(25 - 6) = 5 \times 19.
Answer:9595.
Practice this conceptself-check

Try it yourself

If α,β\alpha,\beta are the roots of x24x+1=0x^2 - 4x + 1 = 0, form the equation whose roots are α2,β2\alpha^2, \beta^2.

From the bank · past-year question

Example 2Quadratic EquationsMODERATE
If pp and qq are the roots of the equation x230x+221=0x^{2} - 30x + 221 = 0, what is the value of p3+q3p^{3} + q^{3} ?

[Q105 · Sep · 2019]

Difference of roots uses −4p, sum of squares uses −2p

(αβ)2=s24p(\alpha-\beta)^2 = s^2 - 4p but α2+β2=s22p\alpha^2+\beta^2 = s^2 - 2p. Mixing the 22 and the 44 is the single most common Vieta error.

Concept 3 of 7

Means of the Roots & Equal-Magnitude Conditions

Intuition

The arithmetic, geometric and harmonic means of the two roots are themselves just sum and product in disguise — so questions about the AM/GM/HM of the roots, or about roots equal in magnitude but opposite in sign, are pure Vieta.

Definition

For roots α,β\alpha, \beta of ax2+bx+c=0ax^2+bx+c=0:

  • AM =α+β2=b2a= \dfrac{\alpha+\beta}{2} = -\dfrac{b}{2a}
  • GM =αβ=ca= \sqrt{\alpha\beta} = \sqrt{\dfrac{c}{a}}
  • HM =2αβα+β=2cb=2cb= \dfrac{2\alpha\beta}{\alpha+\beta} = \dfrac{2c}{-b} = -\dfrac{2c}{b}
  • Roots of equal magnitude, opposite sign: α+β=0\alpha+\beta = 0 (so b=0b = 0) and αβ<0\alpha\beta < 0 (so ca<0\frac{c}{a} < 0). Both conditions are required.

Means of the roots

AM=b2a,GM=ca,HM=2cb\text{AM} = -\tfrac{b}{2a},\quad \text{GM} = \sqrt{\tfrac{c}{a}},\quad \text{HM} = -\tfrac{2c}{b}

Worked example

Find the harmonic mean of the roots of 2x29x+6=02x^2 - 9x + 6 = 0.
  1. Sum =92= \dfrac{9}{2}, product =62=3= \dfrac{6}{2} = 3.
  2. HM =2×productsum=2×39/2=69/2=129= \dfrac{2 \times \text{product}}{\text{sum}} = \dfrac{2 \times 3}{9/2} = \dfrac{6}{9/2} = \dfrac{12}{9}.
Answer:43\dfrac{4}{3}.

From the bank · past-year question

Example 3Quadratic EquationsHARD
The quadratic equation 3x2(k2+5k)x+3k25k=03x^{2}-(k^{2}+5k)x+3k^{2}-5k=0 has real roots of equal magnitude and opposite sign. Which one of the following is correct?

[Q6 · Sep · 2021]

Equal magnitude opposite sign needs TWO conditions

b=0b = 0 alone only makes the roots negatives of each other IF they are real — you also need ca<0\frac{c}{a} < 0 for the roots to be real (and nonzero). A parameter value giving b=0b=0 but ca>0\frac{c}{a}>0 yields imaginary roots, not ±k\pm k.

Concept 4 of 7

Cross-Equation and Shared-Ratio Conditions

Intuition

When a value is a root of one equation and another value is a root of a second, substitute each into its equation and combine the two resulting relations — usually subtracting them is the key move. Two equations sharing a root-ratio link their coefficients.

Definition

Two recurring cross-equation setups:

  • Swapped roots: "nn is a root of x2+px+m=0x^2+px+m=0 and mm is a root of x2+px+n=0x^2+px+n=0" — substitute to get n2+pn+m=0n^2+pn+m=0 and m2+pm+n=0m^2+pm+n=0; subtract them and factor out (nm)(n-m) to get m+n+p=1m+n+p = 1 (when mnm \neq n).
  • Same ratio of roots: if ax2+bx+c=0ax^2+bx+c=0 and px2+qx+r=0px^2+qx+r=0 have roots in the same ratio, then b2ac=q2pr\dfrac{b^2}{ac} = \dfrac{q^2}{pr} (equivalently b2pr=q2acb^2 pr = q^2 ac).

Subtract the substituted equations

n2+pn+m=0,  m2+pm+n=0  (nm)(n+m+p1)=0n^2+pn+m = 0,\ \ m^2+pm+n = 0 \ \Rightarrow\ (n-m)(n+m+p-1) = 0

Worked example

aa is a root of x2+3x+b=0x^2 + 3x + b = 0 and bb is a root of x2+3x+a=0x^2 + 3x + a = 0, with aba \neq b. Find a+ba + b.
  1. Substitute: a2+3a+b=0a^2 + 3a + b = 0 and b2+3b+a=0b^2 + 3b + a = 0.
  2. Subtract: (a2b2)+3(ab)+(ba)=0(ab)(a+b+31)=0(a^2 - b^2) + 3(a - b) + (b - a) = 0 \Rightarrow (a-b)(a+b+3-1) = 0.
  3. Since aba \neq b: a+b+2=0a + b + 2 = 0.
Answer:a+b=2a + b = -2.

From the bank · past-year question

Example 4Quadratic EquationsMODERATE
If nn is a root of the equation x2+px+m=0x^2+px+m=0 and mm is a root of the equation x2+px+n=0x^2+px+n=0, where mnm\neq n, then what is the value of p+m+np+m+n?

[Q16 · Sep · 2024]

Subtract, don't add

Adding the two substituted equations keeps a stubborn m2+n2m^2+n^2; subtracting produces the factor (mn)(m-n) you can cancel. When you see a symmetric pair of "X is a root of … , Y is a root of …", subtract.

Concept 5 of 7

Reducing a Symmetric Equation by Substitution

Intuition

An equation that is symmetric about some centre — like (x−1)⁴ + (x−5)⁴ = constant — flattens to a quadratic once you shift the variable to that centre, because the odd powers cancel. Then Vieta on the full polynomial gives the sum of ALL its roots, real and complex.

Definition

For an equation symmetric about x=mx = m (e.g. (xa)4+(xb)4=k(x-a)^4 + (x-b)^4 = k with centre m=a+b2m = \tfrac{a+b}{2}), substitute u=xmu = x - m. The odd powers of uu cancel, leaving an even equation u4+Au2+B=0u^4 + Au^2 + B = 0 — a quadratic in u2u^2. Solve for u2u^2, count the real roots (each positive u2u^2 gives two), and for the sum of all roots use Vieta on the degree-nn polynomial: sum =(coeff of xn1)(coeff of xn)= -\dfrac{(\text{coeff of }x^{n-1})}{(\text{coeff of }x^{n})} — the complex roots are included.

Shift to the centre of symmetry

(xa)4+(xb)4=k,  u=xa+b2  u4+Au2+B=0(x-a)^4 + (x-b)^4 = k,\ \ u = x - \tfrac{a+b}{2} \ \Rightarrow\ u^4 + Au^2 + B = 0

Worked example

How many real roots does (x1)4+(x3)4=16(x-1)^4 + (x-3)^4 = 16 have?
  1. Centre m=2m = 2; let u=x2u = x - 2: (u+1)4+(u1)4=16(u+1)^4 + (u-1)^4 = 16.
  2. Expand: 2(u4+6u2+1)=16u4+6u27=0(u21)(u2+7)=02(u^4 + 6u^2 + 1) = 16 \Rightarrow u^4 + 6u^2 - 7 = 0 \Rightarrow (u^2-1)(u^2+7) = 0.
  3. u2=1u^2 = 1 gives u=±1u = \pm 1 (real); u2=7u^2 = -7 gives no real uu.
Answer:22 real roots (x=1,3x = 1, 3).

From the bank · past-year question

Example 5Quadratic EquationsMODERATE
Consider the equation (1x)4+(5x)4=82(1-x)^4+(5-x)^4=82.
What is the number of real roots of the equation?

[Q31 · Sep · 2023]

"Number of real roots" ≠ "sum of all roots"

After u2=7u^2 = -7 is rejected for real roots, those two complex roots STILL count toward the sum-of-all-roots (via Vieta on the quartic). Read whether the question wants the real-root count or the full Vieta sum.

Concept 6 of 7

Self-Referential Root Conditions

Intuition

Sometimes the roots themselves appear in the coefficients, or the roots must satisfy extra relations like α+β = α²+β². Just write Vieta's two relations and treat them as simultaneous equations in the unknowns — the system pins everything down.

Definition

When the roots feed back into the equation or an extra relation is imposed, set s=α+βs = \alpha+\beta, p=αβp = \alpha\beta and translate every condition into s,ps, p:

  • Coefficients made of the roots: e.g. x2+αxβ=0x^2 + \alpha x - \beta = 0 with roots α,β\alpha, \beta gives α+β=α\alpha+\beta = -\alpha and αβ=β\alpha\beta = -\beta; solve the pair.
  • Extra symmetric relation: e.g. α+β=α2+β2\alpha+\beta = \alpha^2+\beta^2 becomes s=s22ps = s^2 - 2p; combine with a second given relation to find s,ps, p.
  • A relation true for ALL such roots is established by checking it against ss and pp, not by finding the roots numerically.

Translate every condition into s and p

s=α+β,p=αβ  solve the system in s,ps = \alpha+\beta,\quad p = \alpha\beta \ \Rightarrow\ \text{solve the system in } s, p

Worked example

If α,β\alpha, \beta are the roots of x2+αx+β=0x^2 + \alpha x + \beta = 0 with β0\beta \neq 0, find α\alpha and β\beta.
  1. Vieta: α+β=α\alpha + \beta = -\alpha and αβ=β\alpha\beta = \beta.
  2. From the product (with β0\beta \neq 0): α=1\alpha = 1. Then 1+β=1β=21 + \beta = -1 \Rightarrow \beta = -2.
Answer:α=1, β=2\alpha = 1,\ \beta = -2.

From the bank · past-year question

Example 6Quadratic EquationsHARD
If α\alpha and β\beta (0)(\neq 0) are the roots of the quadratic equation x2+αxβ=0x^2 + \alpha x - \beta = 0, then the quadratic expression x2+αx+β-x^2 + \alpha x + \beta where xRx \in \mathbb{R} has

[Q10 · Sep · 2018]

Don't divide away a root you still need

Cancelling β\beta from αβ=β\alpha\beta = \beta is valid only because β0\beta \neq 0 is given; the discarded case β=0\beta = 0 must be checked separately (it usually fails another condition).

Concept 7 of 7

Structural and Counting Root Problems

Intuition

Some questions are about the STRUCTURE the roots must have — integer coefficients forcing conjugate roots, or an equation unchanged when its roots are squared. These are solved by Vieta plus a constraint argument, not by computation.

Definition

Two structural patterns:

  • Integer / rational coefficients force partner roots. A polynomial with integer coefficients that has a root rr and is of higher degree often forces a partner (e.g. for a quartic with given roots 2,3-2, 3, the symmetric construction (x24)(x29)(x^2-4)(x^2-9) supplies ±2,±3\pm 2, \pm 3).
  • Equation unchanged under a root transform. "The equation is the same when its roots α,β\alpha,\beta are replaced by α2,β2\alpha^2,\beta^2" means {α2,β2}={α,β}\{\alpha^2,\beta^2\} = \{\alpha,\beta\}; enumerate cases (α2=α\alpha^2=\alpha or α2=β\alpha^2=\beta) to count the valid equations. Roots land in {0,1,ω,ω2}\{0, 1, \omega, \omega^2\}.

Unchanged under squaring the roots

{α2,β2}={α,β}  α,β{0,1,ω,ω2}\{\alpha^2, \beta^2\} = \{\alpha, \beta\} \ \Rightarrow\ \alpha, \beta \in \{0, 1, \omega, \omega^2\}

Worked example

A monic quadratic with roots α,β\alpha, \beta is unchanged when the roots are replaced by α2,β2\alpha^2, \beta^2, with α=β\alpha = \beta. What are the possible repeated roots?
  1. Unchanged means α2=α\alpha^2 = \alpha (the repeated root maps to itself).
  2. α2α=0α(α1)=0\alpha^2 - \alpha = 0 \Rightarrow \alpha(\alpha - 1) = 0.
Answer:α=0\alpha = 0 or α=1\alpha = 1.

From the bank · past-year question

Example 7Quadratic EquationsHARD
If 2-2 and 33 are roots of the equation a0+a1x+a2x2+a3x3+x4=0a_0+a_1 x+a_2 x^2+a_3 x^3+x^4=0 where a0,a1,a2,a3a_0,a_1,a_2,a_3 are integers, then which one of the following is correct?

[Q8 · Apr · 2024]

Enumerate the set-equality cases

{α2,β2}={α,β}\{\alpha^2,\beta^2\}=\{\alpha,\beta\} splits into the matched case (α2=α,β2=β\alpha^2=\alpha,\beta^2=\beta) AND the swapped case (α2=β,β2=α\alpha^2=\beta,\beta^2=\alpha). The swapped case is where ω,ω2\omega, \omega^2 enter — miss it and you undercount.

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 (7)

  • Vieta's Relations — Sum and Product of Roots

    Vieta's relations

    α+β=ba,αβ=ca\alpha + \beta = -\dfrac{b}{a}, \qquad \alpha\beta = \dfrac{c}{a}
  • Symmetric Functions & Forming New Equations

    Build the equation from new sum & product

    x2Sx+P=0,S=(new sum), P=(new product)x^2 - S x + P = 0, \quad S = \text{(new sum)},\ P = \text{(new product)}
  • Means of the Roots & Equal-Magnitude Conditions

    Means of the roots

    AM=b2a,GM=ca,HM=2cb\text{AM} = -\tfrac{b}{2a},\quad \text{GM} = \sqrt{\tfrac{c}{a}},\quad \text{HM} = -\tfrac{2c}{b}
  • Cross-Equation and Shared-Ratio Conditions

    Subtract the substituted equations

    n2+pn+m=0,  m2+pm+n=0  (nm)(n+m+p1)=0n^2+pn+m = 0,\ \ m^2+pm+n = 0 \ \Rightarrow\ (n-m)(n+m+p-1) = 0
  • Reducing a Symmetric Equation by Substitution

    Shift to the centre of symmetry

    (xa)4+(xb)4=k,  u=xa+b2  u4+Au2+B=0(x-a)^4 + (x-b)^4 = k,\ \ u = x - \tfrac{a+b}{2} \ \Rightarrow\ u^4 + Au^2 + B = 0
  • Self-Referential Root Conditions

    Translate every condition into s and p

    s=α+β,p=αβ  solve the system in s,ps = \alpha+\beta,\quad p = \alpha\beta \ \Rightarrow\ \text{solve the system in } s, p
  • Structural and Counting Root Problems

    Unchanged under squaring the roots

    {α2,β2}={α,β}  α,β{0,1,ω,ω2}\{\alpha^2, \beta^2\} = \{\alpha, \beta\} \ \Rightarrow\ \alpha, \beta \in \{0, 1, \omega, \omega^2\}

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 EquationsEASY
For how many quadratic equations, the sum of roots is equal to the product of roots?

[Q8 · Sep · 2022]

Example 2Quadratic EquationsMODERATE
What is the value of kk for which the sum of the squares of the roots of 2x22(k2)x(k+1)=02x^{2} - 2(k-2)x - (k+1) = 0 is minimum ?

[Q5 · Sep · 2019]

Example 3Quadratic EquationsHARD
Consider the following for the items that follow: A quadratic equation is given by (3+22)x2(4+23)x+(8+43)=0(3+2\sqrt{2})x^{2}-(4+2\sqrt{3})x+(8+4\sqrt{3})=0.
What is the HM of the roots of the equation?

[Q49 · Apr · 2023]

Example 4Quadratic EquationsMODERATE
If the roots of the equation x2+px+q=0x^2 + px + q = 0 are in the same ratio as those of the equation x2+lx+m=0x^2 + lx + m = 0, then which one of the following is correct?

[Q5 · Apr · 2017]

Example 5Quadratic EquationsMODERATE
Consider the equation (1x)4+(5x)4=82(1-x)^4+(5-x)^4=82.
What is the sum of all the roots of the equation?

[Q32 · Sep · 2023]

Drill every past-year question on this subtopic

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