NIMCET, GATE, CUET & CBSE test series are live — start practicing free
syllabuzAI

Symmetric Functions and Location of Roots | NIMCET

Learn to evaluate symmetric functions of roots such as α² + β² and α³ + β³, transform equations, and use the conditions for roots lying on either side of or between given numbers, for NIMCET.

5 min read

SP

SHIVAM PRADHAN

Updated 25 October 2026

On this page

Quick answer: A symmetric function of the roots α, β is unchanged when α and β are swapped, so it can be written using α+β=−ba\alpha + \beta = -\frac{b}{a} and αβ=ca\alpha\beta = \frac{c}{a}. For example, α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta and α3+β3=(α+β)3−3αβ(α+β)\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta). For location of roots of f(x)=ax2+bx+cf(x) = ax^2 + bx + c with a>0a > 0: a number k lies between the roots exactly when f(k)<0f(k) < 0.

This builds on quadratic equations: roots and discriminant.

Standard symmetric functions

Let S=α+βS = \alpha + \beta and P=αβP = \alpha\beta.

ExpressionIn terms of S and P
α2+β2\alpha^2 + \beta^2S2−2PS^2 - 2P
(α−β)2(\alpha - \beta)^2S2−4PS^2 - 4P
α3+β3\alpha^3 + \beta^3S3−3PSS^3 - 3PS
α4+β4\alpha^4 + \beta^4(S2−2P)2−2P2(S^2 - 2P)^2 - 2P^2
1α+1β\frac{1}{\alpha} + \frac{1}{\beta}SP\frac{S}{P}
αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}S2−2PP\frac{S^2 - 2P}{P}
α2β+αβ2\alpha^2\beta + \alpha\beta^2PSPS

Transforming an equation

If α, β are the roots of f(x)=0f(x) = 0:

New rootsNew equation
α+k\alpha + k, β+k\beta + kf(x−k)=0f(x - k) = 0
kαk\alpha, kβk\betaf(xk)=0f\left(\frac{x}{k}\right) = 0
1α\frac{1}{\alpha}, 1β\frac{1}{\beta}f(1x)=0f\left(\frac{1}{x}\right) = 0, that is, cx2+bx+a=0cx^2 + bx + a = 0
−α-\alpha, −β-\betaf(−x)=0f(-x) = 0

Cubic equations

For ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0 with roots α, β, γ:

RelationValue
α+β+γ\alpha + \beta + \gamma−ba-\frac{b}{a}
αβ+βγ+γα\alpha\beta + \beta\gamma + \gamma\alphaca\frac{c}{a}
αβγ\alpha\beta\gamma−da-\frac{d}{a}

Location of roots

Let f(x)=ax2+bx+cf(x) = ax^2 + bx + c with a>0a > 0, discriminant D and vertex at −b2a-\frac{b}{2a}.

RequirementConditions
k lies between the rootsf(k)<0f(k) < 0
Both roots greater than kD≥0D \ge 0, f(k)>0f(k) > 0, −b2a>k-\frac{b}{2a} > k
Both roots less than kD≥0D \ge 0, f(k)>0f(k) > 0, −b2a<k-\frac{b}{2a} < k
Both roots between k1k_1 and k2k_2D≥0D \ge 0, f(k1)>0f(k_1) > 0, f(k2)>0f(k_2) > 0, k1<−b2a<k2k_1 < -\frac{b}{2a} < k_2
Exactly one root in (k1,k2)(k_1, k_2)f(k1)f(k2)<0f(k_1)f(k_2) < 0

For a<0a < 0, apply the same rules to −f(x)-f(x), or use af(k)<0af(k) < 0 for the first row.

Worked examples

Example 1: α² + β²

If α, β are the roots of x2−5x+3=0x^2 - 5x + 3 = 0, find α2+β2\alpha^2 + \beta^2 and α3+β3\alpha^3 + \beta^3.

Step 1: S=5S = 5, P=3P = 3.

Step 2: α2+β2=25−6=19\alpha^2 + \beta^2 = 25 - 6 = 19.

Step 3: α3+β3=125−45=80\alpha^3 + \beta^3 = 125 - 45 = 80.

Example 2: new equation

Form the equation whose roots are the squares of the roots of x2−5x+3=0x^2 - 5x + 3 = 0.

New sum =19= 19, new product =P2=9= P^2 = 9, so x2−19x+9=0x^2 - 19x + 9 = 0.

Example 3: k between the roots

For what m does 2 lie between the roots of x2−mx+1=0x^2 - mx + 1 = 0?

f(2)=4−2m+1<0f(2) = 4 - 2m + 1 < 0, so m>52m > \frac{5}{2}.

Example 4: both roots greater than 1

Find k so that both roots of x2−2kx+k2−1=0x^2 - 2kx + k^2 - 1 = 0 exceed 1.

Step 1: D=4k2−4(k2−1)=4>0D = 4k^2 - 4(k^2 - 1) = 4 > 0 always.

Step 2: f(1)=1−2k+k2−1=k2−2k>0f(1) = 1 - 2k + k^2 - 1 = k^2 - 2k > 0, so k<0k < 0 or k>2k > 2.

Step 3: Vertex k>1k > 1.

Answer: k>2k > 2. (Indeed the roots are k±1k \pm 1, and k−1>1k - 1 > 1.)

Example 5: cubic relation

If α, β, γ are the roots of x3−6x2+11x−6=0x^3 - 6x^2 + 11x - 6 = 0, find α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2.

(α+β+γ)2−2(αβ+βγ+γα)=36−22=14(\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 36 - 22 = 14.

Summary table

IdeaKey result
α2+β2\alpha^2 + \beta^2S2−2PS^2 - 2P
α3+β3\alpha^3 + \beta^3S3−3PSS^3 - 3PS
Reciprocal rootsSwap a and c
k between roots (a>0a > 0)f(k)<0f(k) < 0
One root in an intervalf(k1)f(k2)<0f(k_1)f(k_2) < 0

Exam tips

  • Rewrite every expression in S and P before substituting.
  • For location problems, think of the parabola's graph: sign of f(k)f(k), position of the vertex, and D≥0D \ge 0.
  • A quick check: find the roots directly if they are simple, as in Example 4.

Next, see arithmetic progression formulas, or browse the NIMCET Algebra chapter.

Common mistakes

  • Writing α2+β2=S2\alpha^2 + \beta^2 = S^2.
  • Forgetting the vertex condition in "both roots greater than k".
  • Using αβγ=da\alpha\beta\gamma = \frac{d}{a} for a cubic; it is −da-\frac{d}{a}.
  • Ignoring the sign of a in location-of-roots conditions.

Frequently Asked Questions

What is α² + β² in terms of the coefficients?

α² + β² = (α + β)² − 2αβ = (b² − 2ac)/a².

What is a symmetric function of roots?

An expression in the roots that does not change when the roots are interchanged, such as α + β or α²β + αβ².

How do you form an equation with reciprocal roots?

Replace x by 1/x. For ax² + bx + c = 0, the new equation is cx² + bx + a = 0.

When does a number k lie between the roots of a quadratic?

For ax² + bx + c with a > 0, when f(k) < 0. In general, when a·f(k) < 0.

What is the product of the roots of a cubic equation?

For ax³ + bx² + cx + d = 0, the product of the roots is −d/a.

Concept Check

MEDIUM

What is the principal value branch (range) of cos⁡−1x\cos^{-1} x?

Suggested for you