Question 125

If $$(n^2-tn+\frac{1}{4})$$ be a perfect square then the value of t are:

Solution

If $$(n^2-tn+\frac{1}{4})$$ need to be a perfect square then ,

$$(n^2-tn+\frac{1}{4})$$ = $$(n^2 + \frac{1}{\pm2}^2 - 2\times\frac{1}{\pm2}n)$$

and hence we can conclude that "t" can take ±1 value

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