Question

If \( \alpha, \beta, \gamma \) are the roots of the equation \( x^3 - Px^2 + Qx - R = 0 \) and \( (\alpha-2)^2, (\beta-2)^2, (\gamma-2)^2 \) are the roots of the equation \( x^3 - 5x^2 + 4x = 0 \), then the possible least value of \( P+Q+R \) is

• A. 5
• B. -7
• C. -1
• D. 1

Hint:

When dealing with symmetric functions of roots (like coefficients of a polynomial), if the roots are non-negative, minimizing the roots individually will minimize their sums and products. Always check the sign of the constant term in the cubic expansion: \( (x-\alpha)(x-\beta)(x-\gamma) = x^3 - (\sum \alpha)x^2 + (\sum \alpha\beta)x - (\alpha\beta\gamma) \).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given two cubic equations. The roots of the second equation are related to the roots of the first. We first find the explicit numerical roots of the second equation and then use them to determine the possible values for \( \alpha, \beta, \gamma \). Finally, we minimize the sum \( P+Q+R \) based on the relations between coefficients and roots.
Step 2: Computation Steps:
The second equation is: \[ x^3 - 5x^2 + 4x = 0 \] Factorizing it: \[ x(x^2 - 5x + 4) = 0 \] \[ x(x-1)(x-4) = 0 \] So, the roots are \( 0, 1, 4 \). We are given that the roots of this equation are \( (\alpha-2)^2, (\beta-2)^2, (\gamma-2)^2 \). Therefore, the set \( \{ (\alpha-2)^2, (\beta-2)^2, (\gamma-2)^2 \} \) is equal to the set \( \{ 0, 1, 4 \} \). We can assign these values to find the possible values for \( \alpha, \beta, \gamma \): 1. \( (\alpha-2)^2 = 0 \implies \alpha - 2 = 0 \implies \alpha = 2 \). 2. \( (\beta-2)^2 = 1 \implies \beta - 2 = \pm 1 \implies \beta = 3 \) or \( \beta = 1 \). 3. \( (\gamma-2)^2 = 4 \implies \gamma - 2 = \pm 2 \implies \gamma = 4 \) or \( \gamma = 0 \). The first equation is \( x^3 - Px^2 + Qx - R = 0 \). Using the relationships between roots and coefficients: \[ P = \alpha + \beta + \gamma \] \[ Q = \alpha\beta + \beta\gamma + \gamma\alpha \] \[ R = \alpha\beta\gamma \] We need to find the least value of \( P+Q+R \). Since \( \alpha, \beta, \gamma \) are all non-negative (from the possible values \( \{2\}, \{1,3\}, \{0,4\} \)), the sum \( P+Q+R \) will be minimized when the roots themselves are as small as possible. Let's choose the minimum possible values for the roots: \( \alpha = 2 \) (fixed) \( \beta = 1 \) (choosing 1 over 3) \( \gamma = 0 \) (choosing 0 over 4) Now Calculation Process \( P, Q, R \) for the set \( \{2, 1, 0\} \): \[ P = 2 + 1 + 0 = 3 \] \[ Q = (2)(1) + (1)(0) + (0)(2) = 2 \] \[ R = (2)(1)(0) = 0 \] The sum is: \[ P + Q + R = 3 + 2 + 0 = 5 \] (For comparison, if we took the maximum roots \( \{2, 3, 4\} \), the sum would be much larger).
Step 4: Required Answer:
The possible least value is 5.