find two whole numbers that their product is 72 and their sum is 38?

Answer 1

xy=72
x+y=38
x=38-y

so (38-y)y = 72
38y-y^2=72
y^2-38y+72=0

(y-36)(y-2)=0
so y= 36 or 2
so x = 2 0r 36
so the numbers are 2 and 36

Answer 2

36 x 2 = 72 and 36 + 2 = 38 so 36 and 2

Answer 3

Let a be the first number, so the second number is 38 - a. tTeir product is 72, so we have the equation

a(38 - a) = 72
38a - a^2 = 72
a^2 - 38a + 72 = 0
(a - 36)(a - 2) = 0
a = 36, 2
38 - a = 2, 36

Therefore, the numbers are 36 and 2.

Answer 4

The numbers are the solutions of : x^2-38x+72=0

--> x= [19±√(19^2-72)] = 19±√289 = 19±17 = 36 and 2

Saludos.

Answer 5

Let nos. be x and y
xy = 72
x + y = 38

x + 72/x = 38
x² + 72 = 38 x
x² - 38 x + 72 = 0
(x - 36) (x - 2) = 0
x = 36 , x = 2
Numbers are 2 and 36

Answer 6

A * B = 72
A + B = 38

... A = 38 - B
... (38 - B) * B = 72
... 38B - B^2 = 72
... 0 = B^2 - 38B + 72
... (B - 36) * (B - 2) = 0
B = 36 or 2
So, the numbers are 36 and 2

Answer 7

let the 2 whole numbers be x and y
ATQ
sum of numbers is 38
=>x+y=38
also
product is 72
=>x*y=72
from first equn
x=y-38
substituting in second equation
(y-38)y=72
y²-38y-72=0
y²-36y-2y-72=0
y(y-36)+2(y-36)=0
(y-2)(y-36)=0
so the numbers are 36 and 2

Answer 8

numbers = x and y
xy=72
x+y=38
x+72/x=38 (xy=72, so y=72/x)
x^2+72-38x=0
x^2-38x+72=0
x=36 or 2
if x =36, y=72/36=2
if x=2, y=72/2=36
so, the numbers are 2 and 36

Answer 9

36 and 2

you dont have to make complicated equations...
all you have to do is list the factors of 72:

36 and 2
18 and 4
9 and 8
3 and 24
1 and 72

then you try to find which of them will have a sum of 38

and so we can see 36 and 2

PS:of course if you already saw which factors has the sum needed even though you haven't finished listing them... you don't need to finish the list.. which would yield a very short method

Answer 10

xy=72
x+y=38
x=38-y

xy=72
(38-y)y=72
38y-y^2=72
y^2-38y+72=0
(y-36)(y-2)=0
y=36 or y=2

x=38-y
x=38-36
x=2

x=38-y
x=38-2
x=36

x=36,y=2 or x=2,y=36