Question

Let alpha is not equals to 1 be a real root of the equation x3 - ax2 + ax – 1=0 where a is not equals to 1 is a real number.
Then a root of this equation among the following is.
1)alpha ^2 2)-1/alpha 3)1/alpha 4)1/alpha^2​

Answer 1

Given :  x³ - ax² + ax - 1  =0

α is real root and not equal to 1

To Find : a root of this equation

Solution:

x³ - ax² + ax - 1  =0

=> x³ - 1   - ax(x  - 1) = 0

=> (x - 1)(x² + x + 1) - ax(x - 1) = 0

=> (x - 1)(x² + x + 1 - ax) = 0

=> x = 1

or

x² + x + 1 - ax = 0

=> x² + x(1 - a) + 1 = 0

α is real root

=> D ≥ 0

(1 - a)² - 4(1)(1)  ≥ 0

=> a² - 2a + 1 - 4   ≥ 0

=> a²  - 2a  - 3   ≥ 0

=> (a - 3)(a + 1) ≥ 0

=> a ≥ 3  or a ≤ - 1

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