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Write two numbers which differ by 3 and whose product is 54.
Given :
Difference between two numbers $= 3$.
Product of those numbers $= 54$.
To do :
We have to find the numbers.
Solution :
Let the one number be $= a$
So, second number $= a - 3$
Now,
$a \times (a - 3) = 54$
$a^2 - 3a = 54$
$a^2 - 3a - 54 = 0$
$a^2 - 9a + 6a - 54 = 0$
$a(a-9) + 6(a - 9) = 0$
$(a+6)(a-9) = 0$
So,
$a = 9$ or $-6$
But a could not be negative so $a =9$.
1st number $= a = 9$
2nd number $= a - 3 = 9 - 3 = 6$
Therefore, the numbers are 9 and 6.Advertisements