two numbers differ by 3 and product is 504 , find the number
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Step-by-step explanation:
Let the two natural numbers be x and y.
Given that difference of numbers are 3.
x−y=3
x=3+y⟶(1)
Also given that product of numbers is 504.
xy=504
From eq
n
(1), we have
⇒ (3+y)y=504
⇒3y+y
2
=504
⇒y
2
+3y−504=0
⇒(y+24)(y−21)=0
⇒y=21 or −24
Case I:-
y=21
Substituting the value of y in eq
n
(1), we have
x=3+y=3+21=24
Hence, x=24.
Case II:-
y=−24
Substituting the value of y in eq
n
(1), we have
x=3+y=3+(−24)=−21
Hence, x=−21.
Hence, the numbers can be (21 and 24) or (-21 and -24).
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