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The sum of two numbers is 48 and its product is 432. Find the numbers.

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Hint: Assume the condition given with variables and proceed the simplification part in a proper manner.

Let us consider x and y are two required numbers.
Given sum of two numbers as 48
$ \Rightarrow x + y = 48 \to (1)$
And also given that product of two numbers as 432
$ \Rightarrow xy = 432 \to (2)$
From equation (2) we can write
$ \Rightarrow y = \dfrac{{432}}{x} \to (3)$
Now on substituting y value in equation (1) we get
$\
   \Rightarrow x + \dfrac{{432}}{x} = 48 \\
   \Rightarrow {x^2} + 432 = 48x \\
   \Rightarrow {x^2} - 48x + 432 = 0 \\
   \Rightarrow {x^2} - 36x - 12x + 432 = 0 \\
   \Rightarrow x(x - 36) - 12(x - 36) = 0 \\
   \Rightarrow (x - 36)(x - 12) = 0 \\
   \Rightarrow x = 36,12 \\
\ $
Now to get y value let us substitute x value in equation (3)
If x=36, $y = \dfrac{{432}}{{36}} = 12$
If x=12, $y = \dfrac{{432}}{{12}} = 36$

Here if we put x=36 we get y=12 and if x=12 then y=36
Therefore we got both x, y values.

Note: Here we have considered two numbers as x and y and applied the given conditions. Later we have substituted x value in one condition to get y value (we can substitute if any of required condition) And on further simplification we get x, y values. Concentrate on factorization of the equation to get proper values.