যদি x2-√5x+1=0 হয় ,তাহলে x2-(1/x)2=?
Solution
Correct Answer: Option B
দেওয়া আছে, x2-√5x+1=0
অতএব, x2 - 1/x2 = (x+1/x) (x-1/x) = √5 × 1 = √5 [ a2-b2 =(a+b)(a-b)]
=> x2 + 1 = x√5
=> (x2 + 1) /x = √5 [x দ্বারা ভাগ করে ]
=> x + (1/x) = √5
এখন,
(x + 1/x)2 = (x - 1/x)2 + 4× x . 1/x
(x + 1/x)2 = (x - 1/x)2 + 4× x . 1/x
=> (√5)2 = (x - 1/x)2 + 4
=> (x - 1/x)2 = 5 - 4 = 1
=> (x - 1/x) = √1 = 1
অতএব, x2 - 1/x2 = (x+1/x) (x-1/x) = √5 × 1 = √5 [ a2-b2 =(a+b)(a-b)]
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