Let p and q be two real numbers such that p + q = 3 and p⁴ + q⁴ = 369. Then (1/p + 1/q)⁻² is equal t

Q: Wait, why can't I just solve for $p$ and $q$ using the quadratic formula immediately?

You absolutely could, but think about the complexity! You would be dealing with nested square roots and potentially complex numbers. In JEE Advanced, the goal is to find the 'algebraic shortcut'—using symmetric polynomials to bypass the need for individual values.

Q: Why did we reject $pq = 24$? Is there a mathematical reason or just a guess?

It is not a guess! If $pq = 24$, then $p^2 + q^2 = (p+q)^2 - 2pq = 9 - 48 = -39$. Since $p$ and $q$ are real, their squares must be non-negative. A sum of squares cannot be negative, so $pq = 24$ is physically impossible in the real number domain.

Q: How do I know which identity to use for $p^4 + q^4$?

Always look at the powers. $p^4 + q^4$ is the square of $(p^2 + q^2)$ minus a correction term. By building up from $(p+q)^2$, you create a chain of dependencies that allows you to isolate the product $pq$.

Q: Is there a trick to factorizing $(pq)^2 - 18pq - 144 = 0$ quickly?

Look for two numbers that multiply to $-144$ and add to $-18$. $-24$ and $+6$ fit perfectly. Practice your quadratic factorization—it is the most common 'bottleneck' in JEE problems!

Q: Does the order of $p$ and $q$ matter?

Not at all! The expression $(\frac{1}{p} + \frac{1}{q})^{-2}$ is symmetric, meaning if you swap $p$ and $q$, the value remains identical. This symmetry is your best friend in algebra.

Question

Let $p$ and $q$ be two real numbers such that $p + q = 3$ and $p^4 + q^4 = 369$. Then $(\frac{1}{p} + \frac{1}{q})^{-2}$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the road to JEE Advanced. Today, we are not just solving an equation; we are peeling back the layers of a beautiful algebraic structure.

Often, when we see variables like $p$ and $q$ raised to the fourth power, our instinct is to panic. However, in the world of competitive mathematics, the most elegant path is rarely the most direct one. We are going to use the power of symmetry to dance around the variables without ever needing to know their exact identities.

### Simplifying the Target

Look at the expression we are chasing: $(\frac{1}{p} + \frac{1}{q})^{-2}$. It looks intimidating, but let's simplify the interior first.

By finding a common denominator, we get:

$$\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq}$$

Now, applying that negative exponent, the expression transforms into $(\frac{pq}{p+q})^2$. Since we already know $p+q = 3$, our entire problem collapses into finding the value of:

$$\frac{(pq)^2}{9}$$

We have successfully reduced a complex problem into a single, focused mission: find the product $pq$.

### Building the Bridge

How do we connect the sum $p+q$ to the product $pq$ when we are given $p^4 + q^4 = 369$? We use the fundamental building blocks of algebra.

We know that $(p+q)^2 = p^2 + q^2 + 2pq$. This allows us to express the sum of squares as:

$$p^2 + q^2 = (p+q)^2 - 2pq = 9 - 2pq$$

Now, we climb to the fourth power. Using the identity $p^4 + q^4 = (p^2 + q^2)^2 - 2(pq)^2$, we can substitute our previous result.

We are essentially creating a quadratic equation for the variable $pq$. By substituting $369 = (9 - 2pq)^2 - 2(pq)^2$, we are no longer looking at $p$ and $q$; we are looking at the behavior of their product.

### The Quadratic Crucible

Expanding $(9 - 2pq)^2$ gives us $81 - 36pq + 4(pq)^2$. When we subtract the $2(pq)^2$ term, we are left with:

$$369 = 81 - 36pq + 2(pq)^2$$

Rearranging this into a standard quadratic form, we arrive at $2(pq)^2 - 36pq - 288 = 0$. Dividing by 2, we get the clean, factorable equation:

$$(pq)^2 - 18pq - 144 = 0$$

Factoring this, we find $(pq - 24)(pq + 6) = 0$. This gives us two candidates for our product: $pq = 24$ or $pq = -6$.

### The Reality Check

In JEE Advanced, the 'real number' constraint is not just a formality—it is a filter. If $pq = 24$, then $p^2 + q^2 = 9 - 2(24) = -39$.

For any real numbers $p$ and $q$, the sum of their squares must be non-negative. A negative result is a physical impossibility. Thus, we must reject $pq = 24$ and embrace $pq = -6$.

### The Final Victory

With $pq = -6$ in hand, we return to our simplified target: $\frac{(pq)^2}{9}$.

Substituting our value, we get:

$$\frac{(-6)^2}{9} = \frac{36}{9} = 4$$

See how the complexity vanished? We didn't need to solve for $p$ or $q$. We simply respected the structure of the equations, navigated the constraints of the real number system, and arrived at the truth. The final answer is **4**.

Animated Solution for Mathematics - Quadratic Equations: Let $p$ and $q$ be two real numbers such that $p + q = 3$ and $p^{4} + q^{4} = 369$ . Then $(\frac{1}{p} + \frac{1}{q})^{- 2}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Simplify the Target Expression

Substitute Known Values

Relate $p^{2} + q^{2}$ to Knowns

Expand to Power 4

Expand the Quadratic Equation

Simplify the Equation

Solve the Quadratic for $pq$

Check Real Number Constraint

Final Calculation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

Simplifying the Target

Building the Bridge

The Quadratic Crucible

The Reality Check

The Final Victory

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