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CDS Mathematics · Quadratic Equations

Symmetric Functions of the Roots

Any expression in α and β that does not change when they swap can be written through α + β and αβ, which the coefficients give directly.

All Quadratic Equations notesCDS Mathematics strategy

Why this matters

Seventeen PYQs, the largest page in the chapter. Never solve for the roots: rewrite the asked expression through the sum and product. Three rewrites cover nearly all of them — α² + β², (α − β)², and α³ + β³.

Concept 1 of 3: Squares, reciprocals and products of the roots

The sum and product are the two building blocks. Squares come from squaring the sum; reciprocals from dividing the sum by the product; and a product like (α+1)(β+1)(\alpha + 1)(\beta + 1) from expanding it.

Definition

With S=α+β=−baS = \alpha + \beta = -\dfrac ba and P=αβ=caP = \alpha\beta = \dfrac ca:

  • α2+β2=S2−2P\alpha^2 + \beta^2 = S^2 - 2P and α4+β4=(α2+β2)2−2P2\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2P^2;
  • 1α+1β=SP\dfrac1\alpha + \dfrac1\beta = \dfrac SP and αβ+βα=S2−2PP\dfrac\alpha\beta + \dfrac\beta\alpha = \dfrac{S^2 - 2P}{P};
  • (α+k)(β+k)=P+kS+k2(\alpha + k)(\beta + k) = P + kS + k^2;
  • since aα2+bα+c=0a\alpha^2 + b\alpha + c = 0, aα+b=−cαa\alpha + b = -\dfrac c\alpha.

Sum of squares of the roots

α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

Worked example

If α,β\alpha, \beta are the roots of x2−5x+3=0x^2 - 5x + 3 = 0, find α2+β2\alpha^2 + \beta^2 and 1α+1β\dfrac1\alpha + \dfrac1\beta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2023 · CDS (I) 2023 — Elementary Mathematics · Q21Moderate

Example 1 · Quadratic Equations · Symmetric Functions of the Roots

If α\alpha and β\beta are the roots of the equation x2−7x+1=0x^2 - 7x + 1 = 0, then what is the value of α4+β4\alpha^4 + \beta^4?

A bound that is never reached

When a parameter varies, an expression like 18k−2\dfrac{18}{k} - 2 with k<0k < 0 gets as close to −2-2 as you like but never equals it. Check whether the extreme value is actually attained before calling it the maximum.
Drill 9 more on squares, reciprocals and products of the roots

Concept 2 of 3: The difference of the roots

(α−β)2(\alpha - \beta)^2 is the square of the sum minus four times the product, which is the discriminant divided by a2a^2. So a condition on the difference of the roots is a condition on the discriminant.

Definition

  • (α−β)2=(α+β)2−4αβ=b2−4aca2(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \dfrac{b^2 - 4ac}{a^2}.
  • If the roots differ by dd: b2−4ac=a2d2b^2 - 4ac = a^2d^2.
  • Sum and difference together give the roots: α=S+d2\alpha = \dfrac{S + d}{2}, β=S−d2\beta = \dfrac{S - d}{2}.
  • Move every term to one side before reading the coefficients: in x2−bx+c=5x^2 - bx + c = 5 the constant is c−5c - 5.

Difference of the roots

(α−β)2=(α+β)2−4αβ(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta

Worked example

The roots of x2−9x+k=0x^2 - 9x + k = 0 differ by 33. Find kk.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (I) 2020 — Elementary Mathematics · Q12Moderate

Example 2 · Quadratic Equations · Symmetric Functions of the Roots

If α\alpha and β\beta are the roots of the quadratic equation x2+kx−15=0x^{2} + kx - 15 = 0 such that α−β=8\alpha - \beta = 8, then what is the positive value of kk ?

The difference fixes k only up to sign

k2=49k^2 = 49 gives k=±7k = \pm 7; the question usually says which sign it wants. Squared conditions always lose the sign.
Drill 4 more on the difference of the roots

Concept 3 of 3: Cubes of the roots

Cube the sum and remove the cross terms, exactly as for any two numbers. When a coefficient depends on a parameter, the result is a function of that parameter whose extreme you can then find.

Definition

  • α3+β3=S3−3PS\alpha^3 + \beta^3 = S^3 - 3PS.
  • α3−β3=(α−β)(S2−P)\alpha^3 - \beta^3 = (\alpha - \beta)(S^2 - P).
  • If SS and PP depend on an angle or a parameter, write the answer in one variable and find its least or greatest value on the allowed range.

Sum of cubes of the roots

α3+β3=(α+β)3−3αβ(α+β)\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)

Worked example

If α,β\alpha, \beta are the roots of x2−3x+1=0x^2 - 3x + 1 = 0, find α3+β3\alpha^3 + \beta^3.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2026 · CDS (II) 2026 — Elementary Mathematics · Q16Hard

Example 3 · Quadratic Equations · Symmetric Functions of the Roots

If α\alpha and β\beta are the roots of the equation 2x2−2(n+1)x+(n2+n+1)=02x^2 - 2(n + 1)x + (n^2 + n + 1) = 0, then what is (2α3+2β3)(2\alpha^3 + 2\beta^3) equal to ?

Keep a leading coefficient

For 2x2−…2x^2 - \ldots the product is the constant DIVIDED BY 22. Forgetting to divide is the commonest slip on this page, and it produces an answer twice too large in one term.
Drill 1 more on cubes of the roots

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 (3)

  • Squares, reciprocals and products of the roots

    Sum of squares of the roots

    α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta
  • The difference of the roots

    Difference of the roots

    (α−β)2=(α+β)2−4αβ(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta
  • Cubes of the roots

    Sum of cubes of the roots

    α3+β3=(α+β)3−3αβ(α+β)\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)

Watch out for (3)

Test yourself on Quadratic Equations

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Quadratic Equations chapter testDrill all 17 questions