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Question 1:
If the equation x3 – ax2 + bx – a = 0 has three real roots then the following is true
A. a = 1
B. a ≠ 1
C. b = 1
D. b ≠ l
Feedback The given equation x3 – ax2 + bx – a = 0 can be rewritten as:
x (x2 + b) – a(x2 + 1) = 0
In case b = 1, then the equation becomes (x – a)( x2 + 1) = 0
Now here x2 + 1 = 0 will give imaginary roots.
Hence, if the given equation has three real roots, then b ≠ 1.

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