Two numbers are differ by 3 and their product is 504 find the numbers
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The answer with solution in given in pic...
The numbers are 24 and 21...
OR
-21 and - 24...
As both the values satisfy the equations.
The numbers are 24 and 21...
OR
-21 and - 24...
As both the values satisfy the equations.
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Let one number be x.
Then the other number differ by 3 = x - 3. ------- (1)
Given that their product is 504.
x(x - 3) = 504
x^2 - 3x - 504 = 0
x^2 - 24x + 21x - 504 = 0
x(x - 24) + 21(x - 24) = 0
(x - 24)(x + 21) = 0
x = 24 (or) x= -21.
Therefore the numbers are 24 and - 21.
Substitute the value of x in (1), we get 24-3, -21-3
= 21,-24.
Therefore the numbers are 24,-21 (or) 21,-24.
Hope this helps!
Then the other number differ by 3 = x - 3. ------- (1)
Given that their product is 504.
x(x - 3) = 504
x^2 - 3x - 504 = 0
x^2 - 24x + 21x - 504 = 0
x(x - 24) + 21(x - 24) = 0
(x - 24)(x + 21) = 0
x = 24 (or) x= -21.
Therefore the numbers are 24 and - 21.
Substitute the value of x in (1), we get 24-3, -21-3
= 21,-24.
Therefore the numbers are 24,-21 (or) 21,-24.
Hope this helps!
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