Mathematics
Show that the equation x2 + ax - 1 = 0 has real and distinct roots for all real values of a.
Quadratic Equations
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Answer
Comparing x2 + ax - 1 = 0 with ax2 + bx + c = 0 we get,
a = 1, b = a and c = -1.
We know that,
Discriminant (D) = b2 - 4ac = (a)2 - 4.(1).(-1) = a2 + 4
Since a2 ≥ 0 for all real values of a, we have D = a2 + 4 > 0
The equation has real and distinct roots for all real values of a.
Hence, equation x2 + ax - 1 = 0 has real and distinct roots for all real values of a.
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