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If a, b are real numbers such that x3-ax2 + bx – 6 = 0 has its roots real and positive then minimum value of b is
- a)1
- b)2
- c)3(36)1/3
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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If a, b are real numbers such that x3-ax2+ bx – 6 = 0 has its ro...
The expression x^3 - ax^2 + bx - a can be factored as (x - a)(x^2 + (b - 1)x + 1).
If a = 1, then the expression becomes x^3 - x^2 + bx - 1 = (x - 1)(x^2 + (b + 1)x + 1).
If a = 2, then the expression becomes x^3 - 2x^2 + bx - 2 = (x - 2)(x^2 + (b - 2)x + 1).
Therefore, if a and b are real numbers such that x^3 - ax^2 + bx - a can be factored as (x - a)(x^2 + (b - 1)x + 1), then a could be either 1 or 2.
If a = 1, then the expression becomes x^3 - x^2 + bx - 1 = (x - 1)(x^2 + (b + 1)x + 1).
If a = 2, then the expression becomes x^3 - 2x^2 + bx - 2 = (x - 2)(x^2 + (b - 2)x + 1).
Therefore, if a and b are real numbers such that x^3 - ax^2 + bx - a can be factored as (x - a)(x^2 + (b - 1)x + 1), then a could be either 1 or 2.
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