Let p and q be two positive number such that p + q = 2 and p⁴ + q⁴ = 272. Then p and q are roots of

Q: Wait, why do we reject the negative root for $k$?

Great catch! The problem explicitly states that $p$ and $q$ are positive numbers. Since $k$ is defined as the product $pq$, and the product of two positive numbers must be positive, we must reject any negative value for $k$. Always check your constraints against the physical or algebraic reality of the problem!

Q: Can we just solve for $p$ and $q$ individually using the system of equations?

You could, but it would be incredibly tedious! You would end up with a system of equations that leads to a high-degree polynomial. In JEE Advanced, the goal is often to find the *structure* of the roots (the quadratic equation) rather than the roots themselves. Using symmetric polynomials is the elegant, intended path.

Q: Is there a faster way than squaring the identity twice?

You could use Newton's Sums, which relates the sums of powers of roots to the coefficients of the polynomial. However, for a quadratic, the method of squaring identities is the most fundamental and least prone to memory errors. Stick to the identities you know well!

Question

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

Options:

(A) $x^2 - 2x + 2 = 0$

(B) $x^2 - 2x + 8 = 0$

(C) $x^2 - 2x + 136 = 0$

(D) $x^2 - 2x + 16 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a locked door. You have the key—the sum of two numbers, $p + q = 2$, and a high-power relationship, $p^4 + q^4 = 272$.

You do not need to know the exact values of $p$ and $q$ to unlock the door; you only need to understand the structure that binds them. This is the beauty of symmetric polynomials.

In the world of JEE Advanced, we rarely brute-force our way to the roots. Instead, we build the equation that defines them.

### The Blueprint

Every quadratic equation with roots $p$ and $q$ follows a rigid, elegant structure:

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

We are already halfway there! We know $p + q = 2$. The only missing piece of our puzzle is the product $pq$. If we find $pq$, the equation reveals itself.

### The Ladder of Identities

We cannot jump from $p + q$ to $p^4 + q^4$ in one leap. We must climb the ladder of algebraic identities.

First, we look at the squares. We know that $(p + q)^2 = p^2 + q^2 + 2pq$. Rearranging this, we find:

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

Substituting our known sum, $p + q = 2$, we get $p^2 + q^2 = 4 - 2pq$.

Now, we ascend to the fourth power. We treat $p^2$ and $q^2$ as our new variables. The identity $(p^2 + q^2)^2 = p^4 + q^4 + 2(p^2q^2)$ allows us to bridge the gap.

We know $p^4 + q^4 = 272$. Substituting our expression for $p^2 + q^2$, we get:

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

### The Algebraic Trap

Let us simplify our life by setting $k = pq$. Our equation becomes:

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

Expanding the binomial $(4 - 2k)^2$ gives us $16 - 16k + 4k^2$. Now, our equation is:

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

Combining like terms, we arrive at $2k^2 - 16k + 16 - 272 = 0$, which simplifies to $2k^2 - 16k - 256 = 0$. Dividing by $2$, we get:

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

Factoring this quadratic, we find $(k - 16)(k + 8) = 0$. This gives us two candidates: $k = 16$ or $k = -8$.

But wait! The problem states $p, q > 0$. Therefore, their product $k$ must be positive. We reject $-8$ and embrace $k = 16$.

### The Final Victory

We have our sum, $p + q = 2$, and our product, $pq = 16$. Plugging these into our blueprint $x^2 - (p + q)x + pq = 0$, we arrive at the final, beautiful result:

**$$x^2 - 2x + 16 = 0$$**

You have successfully navigated the complexity of higher-order polynomials by respecting the symmetry of the roots. This is the essence of mathematical mastery.

Animated Solution for Mathematics - Quadratic Equations: Let $p$ and $q$ be two positive number such that $p + q = 2$ and $p^{4} + q^{4} = 272$ . Then $p$ and $q$ are roots of the equation :

Select Answer:

Visualized Solution

Identify the Goal

The Quadratic Formula Structure

Symmetric Polynomials: Squaring

Substitute Known Sum

Stepping up to Fourth Powers

Setting up the Main Equation

Expand the Binomial

Simplify to a Quadratic in $k$

Solve for $k$

Apply Positivity Constraint

Construct the Final Equation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Blueprint

The Ladder of Identities

The Algebraic Trap

The Final Victory

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let p and q be two positive number such that p+q=2 and p4+q4=272. Then p and q are roots of the equation :