quadratic expression

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If $a$ be the number of integral values of $n,(n>11)$ for which the expression $n^2-19n+89$ is a perfect square. Then $a+1 =$

The answer is 1.

1 solution

Pavan Rohit
Dec 17, 2013

$n^{2}-19n+89$ can be written as $(n-9) \times (n-10) -1$ . It's one less than product of two consecutive numbers . It can be written as $a\times (a+1) -1$ which is $a^{2}+a-1$ . Let us assume this as a perfect square whose root is b. In that case, $b^{2}-a^{2}$ becomes a-1. We know that difference between 2 consecutive squares is sum of their roots. So, it should be a minimum of 2a+1. Hence there are 0 solutions

a is a variable in this answer which is not to be confused with the a given in the question

Pavan Rohit - 7 years, 5 months ago

There's another way: (n-9.5)^{2} - k^{2} = 5/4 (n-9.5+k)(n-9.5-k) = (-0.5) (-2.5) or (0.5) (2.5) n=11 is the highest. Other solution is n=8

Naveen Mathew - 7 years, 5 months ago

quadratic expression

The answer is 1.

1 solution

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