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Question
Find the value(s) of 'a' for which the quadratic equation x2 – ax + 1 = 0 has real and equal roots.
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Solution
Given quadratic equation is x2 – ax + 1 = 0
On comparing it with ax2 + bx + c = 0, we get
a = 1, b = –a and c = 1
For real and equal roots, D = 0
i.e. b2 – 4ac = 0
⇒ (–a)2 – 4(1)(1) = 0
⇒ a2 – 4 = 0
⇒ a2 = 4
⇒ a = ±2
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