## anonymous 3 years ago find the value of n that satisfy n^2+25n+19 be a perfect square with n is an even number ?

1. anonymous

@mukushla @sirm3d

2. anonymous

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

3. anonymous

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 ?????

4. anonymous

right :)

5. anonymous

how about ?? k−2m = 3 .... k+2m = 183

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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.. :))))))

10. anonymous

thank u very much, mukushla...

11. anonymous

welcome :)