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anonymous
 4 years ago
find the value of n that satisfy
n^2+25n+19 be a perfect square
with n is an even number ?
anonymous
 4 years ago
find the value of n that satisfy n^2+25n+19 be a perfect square with n is an even number ?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0unfortunately i dont have enough time for thinkin on this :( try this : \[n^2+25n+19=m^2\]discriminant of quadratic must be a perfect square too\[6254(19m^2)=k^2\]rearranging gives\[(k2m)(k+2m)=549=3^2\times 61\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0k−2m = 9 k+2m = 61 i got k =35 and m = 13 therefore n^2+25n+19169=0 or n^2+25n150=0 (n15)(n10)=0 n=15 or n=10 because n even number, so n=10 right ?????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how about ?? k−2m = 3 .... k+2m = 183

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0opsss. i think i was mistake factor our n^2+25n150=0. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0n^2+25n150=0 (n5)(n+30)=0 n=5 (is not an even number) hmmm :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe there is no answer but check other possibilities for m and k

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0k−2m = 3 k+2m = 183 satisfyied for k=93 and m=45 n^2+25n+19=45^2 n^2+25n+192025=0 n^2+25n2006=0 (n34)(n56)=0 n=34 yohohohohohohohoooooo.... i got it hehe.. :))))))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank u very much, mukushla...
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