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If alpha, beta, gamma are the roots of x...

If `alpha, beta, gamma` are the roots of `x^(3) + 2x^(2) - 4x - 3 = 0` then the equation whose roots are `alpha//3, beta//3, gamma//3` is

A

`x^3 +6x^2 - 36 x -81 =0`

B

` 9x^3 +6x^2 -4x -1=0`

C

`9x^3 +6x^2 + 4x +1=0`

D

` x^3 - 6x^2 + 36 x+81 =0`

Text Solution

AI Generated Solution

The correct Answer is:
To find the equation whose roots are \(\frac{\alpha}{3}, \frac{\beta}{3}, \frac{\gamma}{3}\) given that \(\alpha, \beta, \gamma\) are the roots of the polynomial \(x^3 + 2x^2 - 4x - 3 = 0\), we can follow these steps: ### Step 1: Identify the original polynomial The original polynomial is: \[ f(x) = x^3 + 2x^2 - 4x - 3 \] ### Step 2: Substitute \(x\) with \(3x\) To find the new polynomial whose roots are \(\frac{\alpha}{3}, \frac{\beta}{3}, \frac{\gamma}{3}\), we substitute \(x\) with \(3x\) in the original polynomial: \[ f(3x) = (3x)^3 + 2(3x)^2 - 4(3x) - 3 \] ### Step 3: Expand the expression Now, we expand the expression: \[ f(3x) = 27x^3 + 2 \cdot 9x^2 - 12x - 3 \] This simplifies to: \[ f(3x) = 27x^3 + 18x^2 - 12x - 3 \] ### Step 4: Factor out the common term Next, we can factor out the common term (which is 3) from the polynomial: \[ f(3x) = 3(9x^3 + 6x^2 - 4x - 1) \] ### Step 5: Set the polynomial to zero Since we are interested in the roots, we set the polynomial equal to zero: \[ 9x^3 + 6x^2 - 4x - 1 = 0 \] ### Conclusion Thus, the equation whose roots are \(\frac{\alpha}{3}, \frac{\beta}{3}, \frac{\gamma}{3}\) is: \[ 9x^3 + 6x^2 - 4x - 1 = 0 \] ---
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