Question

sum of two numbers is 1000.difference of their square is 25600.Then find the number

Answer 1

the two numbers are 40root(609) or 987.117 and 40root(593) or 974.063

Answer 2

Heya friend,
here is it:

x + y = 1000 ----------------------------- 1 
$x^{2}$ - $y ^{2}$ = 256000 ---------------------------------- 2

by using elimination method in 2, we get,

x + y = 1000 * x
$x^{2}$ - $y ^{2}$ = 256000 * 1

= $x^{2}$ + xy = 1000x
  $x^{2}$ - $y^{2}$ = 256000

by changing the signs accordingly and by canceling further, we get,

xy + $y^{2}$ = 1000x - 256000
= y (x + y) = 1000x - 256000
or, x + y = 1000x = 256000 / y -------------------------------- 3 

on using 3 in 1, we get,

1000x - 256000 / y = 1000
= 1000x - 256000 = 1000y
= 1000x - 1000y = 256000
= 1000 (x-y) = 256000
=  x - y = 256000 / 1000
= x - y = 256

or, x = 256 + y ---------------------------------------- 4

By using 4 in 1, we get

256 + y + y = 1000
256 + 2y     = 1000
2y                = 1000 - 256
2y = 744
y = 372

and, as x + y = 1000 from 1, we get ,


x + 372 = 1000
x = 1000 - 372
x = 628


so, x = 628 and y = 372