Let n be a positive integer such that n^2 +19n +48 is a perfect square. find the value of n.
Answers
Answered by
7
Hey
Here is your answer,
n^2 + 19n + 48 = 0
n^2 + 3n + 16n + 48 = 0
n ( n+3) + 16 ( n+3) = 0
(n+3) (n+16) = 0
n+3=0
n=-3
n+16=0
n=-16
Hope it helps you!
Here is your answer,
n^2 + 19n + 48 = 0
n^2 + 3n + 16n + 48 = 0
n ( n+3) + 16 ( n+3) = 0
(n+3) (n+16) = 0
n+3=0
n=-3
n+16=0
n=-16
Hope it helps you!
Answered by
6
Answer:
Step-by-step explanation: the previous comment is so useless the Question show that n square 2 plus19 times n plus48 "is a perfect square "not zero!!!then that solution is not works on and so useless
this question is also in the ASMO division m2 /2019 ข้อสุดท้ายด้วย
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