Question:
If $\alpha, \beta , \gamma$ are the roots of the equation $x^3 - 3x^2 + 2x - 1 = 0$ then the value of $[(1 - \alpha) (1 -\beta )(1 - \gamma)]$ is
If $\alpha, \beta , \gamma$ are the roots of the equation $x^3 - 3x^2 + 2x - 1 = 0$ then the value of $[(1 - \alpha) (1 -\beta )(1 - \gamma)]$ is
Updated On: May 12, 2024
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The Correct Option is C
Solution and Explanation
We have; $\alpha, \beta , \gamma = 3$,
$\alpha \beta + \beta \gamma +\gamma \alpha = 2$ and $\alpha \beta \gamma = 1 $
Now, $( 1 - \alpha)(1 - \beta )(1- \gamma)$
$ = (1 - \beta - \alpha + \alpha \beta ) (1 - \gamma)$
$ = (1 - \beta - \alpha + \alpha \beta - \gamma + \beta \gamma + \alpha \gamma - \alpha \beta \gamma)$
$ = [1 - ( \alpha + \beta + \gamma ) + ( \alpha \beta + \beta \gamma + \gamma \alpha) - \alpha \beta \gamma]$
$ = [ 1 - 3 + 2 - 1] = - 1$
$\alpha \beta + \beta \gamma +\gamma \alpha = 2$ and $\alpha \beta \gamma = 1 $
Now, $( 1 - \alpha)(1 - \beta )(1- \gamma)$
$ = (1 - \beta - \alpha + \alpha \beta ) (1 - \gamma)$
$ = (1 - \beta - \alpha + \alpha \beta - \gamma + \beta \gamma + \alpha \gamma - \alpha \beta \gamma)$
$ = [1 - ( \alpha + \beta + \gamma ) + ( \alpha \beta + \beta \gamma + \gamma \alpha) - \alpha \beta \gamma]$
$ = [ 1 - 3 + 2 - 1] = - 1$
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
