Solving for the Two Numbers Given Their Sum and Product
Consider the problem where we are given the sum and product of two numbers. Specifically, let's explore a situation where the sum of two numbers is 25 and their product is 50. Our goal is to find these two numbers.
System of Equations
Let the two numbers be ( x ) and ( y ). We can set up the following system of equations based on the provided information:
( x y 25 ) ( xy 50 )Solving the Equations
We can express ( y ) in terms of ( x ) using the first equation:
( y 25 - x )
Next, substitute this expression for ( y ) into the second equation:
( x(25 - x) 50 )
Expanding and rearranging the equation gives us a quadratic equation:
( 25x - x^2 50 )
( x^2 - 25x 50 0 )
To solve this quadratic equation, we use the quadratic formula:
( x frac{-b pm sqrt{b^2 - 4ac}}{2a} )
where ( a 1 ), ( b -25 ), and ( c 50 ).
Plugging in the values, we get:
( x frac{25 pm sqrt{(-25)^2 - 4 cdot 1 cdot 50}}{2 cdot 1} )
( x frac{25 pm sqrt{625 - 200}}{2} )
( x frac{25 pm sqrt{425}}{2} )
( x frac{25 pm 5sqrt{17}}{2} )
Now, we find ( y ) using ( y 25 - x ):
( y 25 - frac{25 pm 5sqrt{17}}{2} )
( y frac{50 - 25 pm 5sqrt{17}}{2} )
( y frac{25 pm 5sqrt{17}}{2} )
Therefore, the two numbers are:
( x frac{25 5sqrt{17}}{2} ) and ( y frac{25 - 5sqrt{17}}{2} )
or vice versa.
Approximate Values
The approximate values of the numbers are:
( x approx 24.29 ) and ( y approx 0.71 )
Verification
A verification can be done using the derived formulas. For example:
( a - b sqrt{ab^2 - 4ab} )
( a - b sqrt{22^2 - 4 cdot 72} )
( a - b sqrt{484 - 288} )
( a - b sqrt{196} )
( a - b 14 )
( a b frac{22 pm 14}{2} )
Thus, ( a 18 ) and ( b 4 ) or ( a 4 ) and ( b 18 ).
Further Examples
For additional problems with similar structures, consider the following:
xy 20 and ( xy 64 )
( y frac{64}{x} )
( x cdot frac{64}{x} 20 )
( x^2 - 264 0 )
( x^2 - 16 - 4x64 0 )
( x(x - 16) 16 )
( x - 4x - 16 0 )
( x 4 )
( y 16 cdot 4 )
( text{Thus, the numbers are: 4 and 16} )
These examples illustrate the application of the problem-solving technique to a variety of scenarios, ensuring a robust understanding of the underlying mathematical concepts.