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tanjung
find the value of n that satisfy n^2+25n+19 be a perfect square with n is an even number ?
unfortunately 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\[625-4(19-m^2)=k^2\]rearranging gives\[(k-2m)(k+2m)=549=3^2\times 61\]
k−2m = 9 k+2m = 61 i got k =35 and m = 13 therefore n^2+25n+19-169=0 or n^2+25n-150=0 (n-15)(n-10)=0 n=15 or n=10 because n even number, so n=10 right ?????
how about ?? k−2m = 3 .... k+2m = 183
opsss. i think i was mistake factor our n^2+25n-150=0. :)
n^2+25n-150=0 (n-5)(n+30)=0 n=5 (is not an even number) hmmm :(
maybe there is no answer but check other possibilities for m and k
k−2m = 3 k+2m = 183 satisfyied for k=93 and m=45 n^2+25n+19=45^2 n^2+25n+19-2025=0 n^2+25n-2006=0 (n-34)(n-56)=0 n=34 yohohohohohohohoooooo.... i got it hehe.. :))))))
thank u very much, mukushla...