If (x-1/x) = 5, the value of x^2+1/x^2 is? (With Full Process Please)
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Answered by
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x-1/x = 5
squaring on both sides
(x-1/x)^2 = 5^2
x^2+(1/x)^2-2*x*1/x = 25
x^2+1/x^2-2 = 25 ( since x and 1/x gets cancelled)
x^2+1/x^2 = 25+2 (by sending 2 on rhs)
x^2+1/x^2 = 27
squaring on both sides
(x-1/x)^2 = 5^2
x^2+(1/x)^2-2*x*1/x = 25
x^2+1/x^2-2 = 25 ( since x and 1/x gets cancelled)
x^2+1/x^2 = 25+2 (by sending 2 on rhs)
x^2+1/x^2 = 27
Answered by
1
x-1/x = 5
squaring on both sides
= (x-1/x)²
= 5² x²+(1/x)²-2*x*1/x
= 25 x²+1/x²-2 = 25
= x²+1/x² = 25+2
= x²+1/x² = 27
squaring on both sides
= (x-1/x)²
= 5² x²+(1/x)²-2*x*1/x
= 25 x²+1/x²-2 = 25
= x²+1/x² = 25+2
= x²+1/x² = 27
Anonymous:
I hope it will help you.
Yes it did.. :) Thnx
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