Let p and q Be Two Positive Numbers Such That p + q = 2 and p⁴ + q⁴ = 272. Then p and q Are Roots of the Equation

JEE Main 2021 — 24 February, Morning Shift. Previous Year Question.

Problem

Let $p$ and $q$ be two positive numbers such that $p+q=2$ and $p^4+q^4=272$. Then $p$ and $q$ are roots of the equation:

Key insight. Every option already agrees on the sum of the roots being $2$ — so the only thing distinguishing them is the product $pq$. Since $p^4+q^4$ can be rewritten purely in terms of $p+q$ and $pq$ (via repeated use of $a^2+b^2=(a+b)^2-2ab$), the given condition becomes a solvable equation for $pq$ alone.

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Approach

Express $p^4+q^4$ in terms of $p+q$ and $pq$ by first finding $p^2+q^2=(p+q)^2-2pq$, then squaring that to get $p^4+q^4=(p^2+q^2)^2-2(pq)^2$. Substitute the given values, solve the resulting equation for $pq$, and use the standard form $x^2-(p+q)x+pq=0$ to write the final quadratic.

Solution

Step 1 — Express p² + q² in terms of pq

Let $s=pq$. Since $p+q=2$:

$$p^2+q^2 = (p+q)^2-2pq = 4-2s$$

Step 2 — Express p⁴ + q⁴ in terms of s

$$p^4+q^4 = (p^2+q^2)^2-2(pq)^2 = (4-2s)^2-2s^2$$

Step 3 — Substitute the given value 272

$$(4-2s)^2-2s^2 = 272$$

$$16-16s+4s^2-2s^2 = 272$$

$$2s^2-16s+16-272 = 0 \implies 2s^2-16s-256=0$$

Step 4 — Simplify and solve for s

Dividing by 2:

$$s^2-8s-128=0$$

$$s = \frac{8\pm\sqrt{64+512}}{2} = \frac{8\pm\sqrt{576}}{2} = \frac{8\pm24}{2}$$

$$s = 16 \quad \text{or} \quad s=-8$$

Step 5 — Reject the invalid value

Since $p,q$ are both positive, their product $pq$ must be positive:

$$pq = 16$$

Step 6 — Write the quadratic with roots p and q

$$x^2-(p+q)x+pq=0 \implies x^2-2x+16=0$$

Answer

Common mistakes

• Forgetting to reject $s=-8$. The quadratic in $s$ naturally has two roots, but since $p,q>0$ forces $pq>0$, only the positive root is valid — using $s=-8$ would give a nonsensical negative product for two positive numbers.
• Mis-expanding $(4-2s)^2$. This expands to $16-16s+4s^2$, not $16-8s+4s^2$ — a slip here changes every subsequent coefficient in the quadratic-in-$s$ equation.

Practise next

• Let $m,n$ be two real numbers with $m+n=3$ and $m^4+n^4=257$. Find the quadratic equation having $m$ and $n$ as its roots, using the same power-sum substitution.