Question

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

Answer 1

@mukushla
@sirm3d

Answer 2

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\[(k-2m)(k+2m)=549=3^2\times 61\]

Answer 3

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 ?????

Answer 4

right :)

Answer 5

how about ?? k−2m = 3 .... k+2m = 183

Answer 6

opsss. i think i was mistake factor our n^2+25n-150=0. :)

Answer 7

n^2+25n-150=0
(n-5)(n+30)=0
n=5 (is not an even number)
hmmm :(

Answer 8

maybe there is no answer but check other possibilities for m and k

Answer 9

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.. :))))))

Answer 10

thank u very much, mukushla...

Answer 11

welcome :)