Math Problem

Ok. this is an "Extreme Challenge" question, according to my son. I have to admit I am not sure on this one. I have solved it, but not correctly apparently lol

Here we go.

Two real numbers such that one number plus 12.5 equals 1/2 the other number. I Take this to mean: x/2=y+12.5 or y/2 = x+12.5

Additionally, the sum of the squares of those numbers have a difference of 200. I take this as:

X squared - Y squared = 200
Y squared-X squared = 200

Solve for X and Y.

The only other statement is that it not need to be in decimal form. An exact fraction is preferred.

My brain hurts.
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You can just stick with y/2 = x+12.5.

>> the sum of the squares of those numbers have a difference of 200
Did you see this problem in official print or verbally or in notes?

"sum of the squares of those numbers" is a common math expression meaning:
(X squared + Y squared)

whereas your expression: X squared - Y squared might be read as difference between two squares.

If you can get a look at the text of the official problem, it would be interesting to see if it matches your statement.
If you really mean "Y squared-X squared = 200" then x= (-50 ± 35)/3
But I see no "sum of the squares" in "Y squared-X squared = 200"
On the other hand, since "the sum of the squares of those numbers" is a single number, I don't understand  with what it would have a difference of 200
But if you meant the sum of the squares of those numbers is 200, then x = -10 ± sqrt(15)
flubbsterAuthor Commented:
ok... Quote " The square of the two  numbers is 200 apart."

My wording was off previously... my apologies.

What the heck does that mean????
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I think it must mean "The squares of the two  numbers are 200 apart."
flubbsterAuthor Commented:
I am so confused. Based on what I just read, I have lost the path I was following to a solution.
y/2 = x+12.5
Y squared-X squared = 200
(2x+25)^2 - x^2 = 200
3x^2 + 100x + 425  = 0

X squared-Y squared = 200
x^2  - (2x+25)^2 = 200
-3x^2 - 100x - 825 = 0
x = (100 ± 10)/-6
Or, if the problem actually does not specify which number is which, then you are right is supposing two possibilities for the difference of two squares. This means that the equations could lead to:

y/2 = x+12.5
(2x+25)^2 - x^2 = -200
Cornelia YoderArtistCommented:
Start with  x/2 = y + 12.5
So  x = 2y + 25

(2y+25)^2 - y^2 = 200
(4y^2 +100y + 625)  -  y^2 = 200
3y^2 + 100y + 425 = 0
y =  -100 +/- Sqrt[100^2 - (4)(3)(425)] / (2)(3)  (Quadratic Formula)
y = -100 +/- Sqrt[4900] / 6
y = (-100 +/- 70) / 6
y = -170/6   or   y = -30/6 = -5

Then x = 2y + 25 (can be left to the reader).

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Cornelia YoderArtistCommented:
OK, I see some confusion about what the problem actually is, so here is a method.

1.   Write the linear equation (no squares in it) so that x = some stuff with y in it.
2.   Write the quadratic equation (the one with squares in it) substituting the "some stuff with y in it" for x.
3.   Multiply out any complex terms, add and subtract like terms, until you have something of the form
      ay^2 + by + c = 0
4.   Apply the Quadratic Formula to solve for y (two solutions).
5.   Apply the linear equation in 1 above, to solve for x.

Quadratic Formula:
What grade is your son in?
flubbsterAuthor Commented:
Junior high. I see where I went wrong. I was trying to come up with two linear equations.
Since this is an "Extreme Challenge", then you approach the problem more extremely. Instead of combining the linear and non-linear equation to come up with a blasé parabola, treat the non-linear equation as a hyperbola, graph it, and then try to draw the line and see if it touches the hyperbola. (I say "try" because some lines may not cross, and that leads to imaginary numbers.)

When he sees the graphs, maybe he'll come up with some repetitious, computative approach on getting the answer without actually solving the problem like others have done above. He could then give a presentation and his interest in math may be sparked (or not).

For your interest, here are two solutions:
x + 12.5 = y/2; x^2 - y^2 = 200

x + 12.5 = y/2; x^2 - y^2 = -200
That's four solutions.
flubbsterAuthor Commented:
Thanks to all. While some helped with exact solution, I found other posts to be interesting and felt it deserved something. Love this community.
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