# If p+q+r =5 and pq+qr+pr = 3 where p,q,r are positive real numbers. Find the largest value which p can take

May 27, 2024 2024-05-27 8:13## If p+q+r =5 and pq+qr+pr = 3 where p,q,r are positive real numbers. Find the largest value which p can take

# If p+q+r =5 and pq+qr+pr = 3 where p,q,r are positive real numbers. Find the largest value which p can take

## Exploring Polynomial Roots: Finding the Largest Value of \( p \)

### Problem Statement:

We are given three positive real numbers \( p \), \( q \), and \( r \) that satisfy the following conditions:

- \( p + q + r = 5 \)
- \( pq + qr + pr = 3 \)

Our goal is to determine the largest possible value for \( p \).

### Step-by-Step Solution:

#### 1. Formulating the Polynomial:

We start by recognizing that the given conditions can be associated with the roots of a cubic polynomial. If \( p, q, \) and \( r \) are roots, the polynomial can be expressed as:

\[ t^3 – (p+q+r)t^2 + (pq + qr + pr)t – pqr = 0 \]

Given:

\( p + q + r = 5 \)

\( pq + qr + pr = 3 \)

We substitute these into the polynomial:

\[ t^3 – 5t^2 + 3t – pqr = 0 \]

#### 2. Reducing the Problem:

To find the maximum value of \( p \), we express \( r \) in terms of \( p \) and \( q \):

\[ r = 5 – (p + q) \]

We substitute \( r \) into the second given condition:

\[ pq + q(5 – (p + q)) + p(5 – (p + q)) = 3 \]

\[ pq + 5q – q^2 – pq + 5p – p^2 – pq = 3 \]

\[ 5p + 5q – q^2 – p^2 – pq = 3 \]

#### 3. Solving the Quadratic Inequality:

Rewriting the equation:

\[ -p^2 + (5 – p)q + (5p – 3) = 0 \]

This is a quadratic in \( q \):

\[ q^2 – (5 – p)q + (p^2 – 5p + 3) = 0 \]

For \( q \) to be real, the discriminant of this quadratic must be non-negative:

\[ \Delta = (5 – p)^2 – 4(p^2 – 5p + 3) \]

\[ = 25 – 10p + p^2 – 4p^2 + 20p – 12 \]

\[ = -3p^2 + 10p + 13 \]

We solve the inequality:

\[ -3p^2 + 10p + 13 \geq 0 \]

#### 4. Finding the Roots of the Discriminant:

We find the roots of \( -3p^2 + 10p + 13 = 0 \):

\[ p = \frac{-10 \pm \sqrt{100 – 4(-3)(13)}}{2(-3)} \]

\[ p = \frac{-10 \pm \sqrt{100 + 156}}{-6} \]

\[ p = \frac{-10 \pm \sqrt{256}}{-6} \]

\[ p = \frac{-10 \pm 16}{-6} \]

This gives us:

\[ p = \frac{6}{-6} = -1 \]

\[ p = \frac{-26}{-6} = \frac{13}{3} \]

So, \( -1 \leq p \leq \frac{13}{3} \). Since \( p \) must be positive, the feasible range for \( p \) is:

\[ 0 < p \leq \frac{13}{3} \]

#### 5. Conclusion:

Given \( p \), \( q \), and \( r \) are positive real numbers, the largest value \( p \) can take is:

\[ p = \frac{13}{3} \]

### Summary:

By translating the given conditions into a polynomial and analyzing its roots, we found that the largest value \( p \) can take is \( \frac{13}{3} \). This elegant solution demonstrates the power of algebraic manipulation and polynomial root analysis in solving seemingly complex problems.