Consider the equation x^2+4x-n=0,where n in [ | vedclass

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(C) The given equation is $x^2+4x-n=0$.Completing the square,we get $x^2+4x+4 = n+4$,which simplifies to $(x+2)^2 = n+4$.For the roots to be integers,

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$6$

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(C) The given equation is $x^2+4x-n=0$.Completing the square,we get $x^2+4x+4 = n+4$,which simplifies to $(x+2)^2 = n+4$.For the roots to be integers,$n+4$ must be a perfect square,say $k^2$,where $k$ is an integer.Given $20 \leq n \leq 100$,we have $24 \leq n+4 \leq 104$.Thus,$24 \leq k^2 \leq 104$.The possible perfect squares $k^2$ in this range are $25, 36, 49, 64, 81, 100$.Correspondingly,$n = k^2 - 4$ gives $n \in \{21, 32, 45, 60, 77, 96\}$.There are $6$ such distinct values of $n$.

Consider the equation $x^2+4x-n=0$,where $n \in [20, 100]$ is a natural number. Then the number of all distinct values of $n$,for which the given equation has integral roots,is equal to

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