SOLUTION: The difference of two positive numbers is 99. The numbers multiply to 630. What is the sum of their squares?
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Question 1209119: The difference of two positive numbers is 99. The numbers multiply to 630. What is the sum of their squares? Found 2 solutions by math_tutor2020, ikleyn:Answer by math_tutor2020(3817) (Show Source):
Let's say x and y are the two numbers where x is the bigger value.
x > y and both are larger than 0
xy = 630 is the product of the values which will be useful later.
x-y = 99 since the two numbers are 99 units apart, ie. their difference is 99
Since your teacher wants to know the sum of the squares, let's square both sides to see what happens.
(x-y)^2 = 99^2
x^2-2xy+y^2 = 9801
x^2+y^2-2xy = 9801
x^2+y^2-2*(xy) = 9801
x^2+y^2-2*(630) = 9801 ..... plug in xy = 630
x^2+y^2-1260 = 9801
x^2+y^2 = 9801+1260
x^2+y^2 = 11061 is the sum of the squares of the two mystery numbers.
Notice how we don't need to find x and y themselves.
However, if you wanted to, you could follow that pathway.
Ignore this section if you prefer the previous method shown above.
x-y = 99 can be rearranged to y = x-99
Plug that into the other equation to get,
xy = 630
x*( y ) = 630
x*(x-99) = 630
x^2-99x = 630
x^2-99x-630 = 0
Then apply the quadratic formula to get the roots x = -6 and x = 105.
I'll let the student handle the scratch work.
We ignore x = -6 because the instructions mention the numbers are positive.
If x = 105 then y = x-99 = 105-99 = 6.
We determine that x = 105 and y = 6 are the two mystery numbers.
The scratch work shown below confirms we have the correct x,y pairing.
x-y = 105-6 = 99
x*y = 105*6 = 630
Then,
x^2+y^2 = 105^2+6^2 = 11061 is the final answer.
The idea of the solution is to present the sum of the squares (the desired value)
via given information by a simplest way, using well known identities.
It is how this problem is designed and how to it is expected to be solved.
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