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JEE Mains Maths · Quadratic Equations

Symmetric Functions and Power Sums

Writing any symmetric expression in the roots — sums of squares, fourth powers, reciprocals — in terms of the sum and product, and cutting high powers down with a recurrence.

All Quadratic Equations notesJEE Mains Maths strategy

Why this matters

Twenty-two PYQs, fifteen of them multiple choice. Nine rebuild expressions like α² + β², α⁴ + β⁴ or 1/α² + 1/β² from the sum and product, often to find an unknown coefficient; thirteen handle high powers such as α²⁵ + β²⁵, mostly with the recurrence the equation itself gives. Two ideas cover the page.

Concept 1 of 2: Symmetric expressions from the sum and product

An expression that stays the same when α\alpha and β\beta swap can always be written in s=α+βs=\alpha+\beta and p=αβp=\alpha\beta. Build it in steps: squares first, then fourth powers from the squares. A sum of reciprocal powers is the same expression divided by a power of pp. When ss or pp is unknown, set the given value equal to the expression and solve for it.

Definition

  • α2+β2=s2−2p\alpha^2+\beta^2=s^2-2p.
  • α3+β3=s3−3ps\alpha^3+\beta^3=s^3-3ps.
  • α4+β4=(s2−2p)2−2p2\alpha^4+\beta^4=(s^2-2p)^2-2p^2.
  • 1α2+1β2=s2−2pp2\frac1{\alpha^2}+\frac1{\beta^2}=\frac{s^2-2p}{p^2}.

Squares to fourth powers

α4+β4=(α2+β2)2−2(αβ)2\alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha\beta)^2

Worked example

The roots of x2−3x+1=0x^2-3x+1=0 are α,β\alpha,\beta. Find α4+β4\alpha^4+\beta^4.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q137Moderate

Example 1 · Quadratic Equations · Symmetric Functions and Power Sums

Let αθ\alpha_{\theta} and βθ\beta_{\theta} be the distinct roots of 2x2+(cos⁡θ)x−1=0,θ∈(0,2π)2x^{2}+ (\cos\theta)x - 1 = 0,\theta \in (0,2\pi). If mm and MM are the minimum and the maximum values of αθ4+βθ4\alpha_{\theta}^{4}+\beta_{\theta}^{4}, then 16(M+m)16(M + m) equals :

Check that the roots can exist

Solving for the product often gives two values. For real roots keep only one with s2−4p≥0s^2-4p\ge0; if the numbers must also be positive, the product must be positive too.
Drill 8 more on symmetric expressions from the sum and product

Concept 2 of 2: High powers by a recurrence

If α\alpha is a root of x2−sx+p=0x^2-sx+p=0, then α2=sα−p\alpha^2=s\alpha-p. Multiply by αn−2\alpha^{n-2}: αn=sαn−1−pαn−2\alpha^n=s\alpha^{n-1}-p\alpha^{n-2}. The same holds for β\beta, so any Pn=kαn+mβnP_n=k\alpha^n+m\beta^n obeys the same rule. A question like P20−3P19P18\frac{P_{20}-3P_{19}}{P_{18}} is built to collapse: match its coefficients to the equation and the ratio is a constant.

Definition

  • x2−sx+p=0x^2-sx+p=0 gives Pn=sPn−1−pPn−2P_n=sP_{n-1}-pP_{n-2}.
  • It holds for αn+βn\alpha^n+\beta^n, αn−βn\alpha^n-\beta^n and any kαn+mβnk\alpha^n+m\beta^n.
  • Match the question's coefficients to the recurrence; the leftover term is what remains.

Power-sum recurrence

x2=sx−p ⇒ Pn=sPn−1−pPn−2x^2=sx-p\ \Rightarrow\ P_n=sP_{n-1}-pP_{n-2}

Worked example

The roots of x2−3x−5=0x^2-3x-5=0 are α,β\alpha,\beta, and Pn=αn+βnP_n=\alpha^n+\beta^n. Find P20−3P19P18\frac{P_{20}-3P_{19}}{P_{18}}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q169Moderate

Example 2 · Quadratic Equations · Symmetric Functions and Power Sums

Let α,β;α>β\alpha,\beta;\alpha > \beta, be the roots of the equation x2−2x−3=0x^{2}-\sqrt{2}x -\sqrt{3}= 0. Let Pn=αn−βn,n∈NP_{n}=\alpha^{n}-\beta^{n},n \in N. Then (113−102)P10+(112+10)P11−11P12(11\sqrt{3}- 10\sqrt{2})P_{10}+ (11\sqrt{2}+ 10)P_{11}- 11P_{12} is equal to:

Read the sign of the product

For x2−sx+p=0x^2-sx+p=0 the rule is Pn=sPn−1−pPn−2P_n=sP_{n-1}-pP_{n-2}. With x2−2x−3=0x^2-2x-3=0, p=−3p=-3, so Pn=2Pn−1+3Pn−2P_n=2P_{n-1}+3P_{n-2}: the minus sign becomes a plus.
Drill 12 more on high powers by a recurrence

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 (2)

  • Symmetric expressions from the sum and product

    Squares to fourth powers

    α4+β4=(α2+β2)2−2(αβ)2\alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha\beta)^2
  • High powers by a recurrence

    Power-sum recurrence

    x2=sx−p ⇒ Pn=sPn−1−pPn−2x^2=sx-p\ \Rightarrow\ P_n=sP_{n-1}-pP_{n-2}

Watch out for (2)

Test yourself on Quadratic Equations

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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