1. If $$(n^2-tn+\frac{1}{4})$$ be a perfect square then the value of t are:
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By: anil on 05 May 2019 03.25 am
If $$(n^2-tn+frac{1}{4})$$ need to be a perfect square then , $$(n^2-tn+frac{1}{4})$$ = $$(n^2 + frac{1}{pm2}^2 - 2 imesfrac{1}{pm2}n)$$ and hence we can conclude that "t" can take ±1 value
By: anil on 05 May 2019 03.25 am
If $$(n^2-tn+frac{1}{4})$$ need to be a perfect square then , $$(n^2-tn+frac{1}{4})$$ = $$(n^2 + frac{1}{pm2}^2 - 2 imesfrac{1}{pm2}n)$$ and hence we can conclude that "t" can take ±1 value
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