Question

the difference of two numbers is 3 and their product is 504. find the numbers

Answer 1

If one number is x then the other number will be x - 3.   ---- (1)

Given that x(x-3) = 504

                  x^2 - 3x - 504 = 0

                  x^2 - 24x + 21x - 504 = 0

                  x(x - 24) + 21(x - 24) = 0

                  x = 24 or - 21.

Substitute x = 24 (or) 21 in (1) we get 24,21 and -24,-21

The numbers are 24,21 (or) -24,-21.         

Answer 2

a-b = 3(given) eq. I.
Squaring both sides ;

(a-b) ^2 = 3^2

a^2 + b^2 - 2ab = 9

Since ab = 504(given)

a^2 + b^2 - 2*504 = 9
a^2 + b^2 - 1008=9
a^2 + b^2 = 9+1008
a^2 + b^2 =1017
(a+b) ^2 - 2ab = 1017
(a+b)^2 - 2*504=1017
(a+b)^2 - 1008=1017
(a+b)^2 =1017+1008
(a+b)^2 = 2025
(a+b)= Under root 2025
(a+b)= 45 eq. II.

Add eq. No. I and II

a-b + a+b = 3+45
2a = 48
a=48/2
a=24

Then b = 24-3
=21

So numbers are 24 and 21.

Hope it's helpful to u.