Determine 2 numbers w/ sum, products, diff. of sqrs equal  TOPIC_SOLVED

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

Determine 2 numbers w/ sum, products, diff. of sqrs equal

Postby Luke53 on Sat May 14, 2011 2:10 pm

Determine two numbers: their sum, their product and the difference of their squares are equal.
Please help.
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Re: Determine 2 numbers w/ sum, products, diff. of sqrs equa

Postby maggiemagnet on Sat May 14, 2011 4:37 pm

Luke53 wrote:Determine two numbers: their sum, their product and the difference of their squares are equal.

Try using variables: x+y = xy = x2 - y2

If you square both sides of x+y = xy, what do you get? Can you use this with the third "half"?

:clap:
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Re: Determine 2 numbers w/ sum, products, diff. of sqrs equa

Postby Luke53 on Sat May 14, 2011 5:19 pm

Luke53 schreef:
Determine two numbers: their sum, their product and the difference of their squares are equal.

Try using variables: x+y = xy = x2 - y2

If you square both sides of x+y = xy, what do you get? Can you use this with the third "half"?

Squaring: (x+y)^2 = x² + y² + 2 x y and (x y)^2 = x² y², I don't see any relation with X² - y², so what to do next please?
Thanks.
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Re: Determine 2 numbers w/ sum, products, diff. of sqrs equa

Postby maggiemagnet on Sat May 14, 2011 7:51 pm

Play around with it. See if some other arrangement, equality, squaring, etc, leads somewhere potentially useful. Try stuff!

:clap:

For instance, if you factor the difference of squares for the third "half", can you substitute from either of the other two "halves"? Can you create a new equation that can factor helpfully? And so forth.
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Re: Determine 2 numbers w/ sum, products, diff. of sqrs equa  TOPIC_SOLVED

Postby Martingale on Sat May 14, 2011 8:55 pm

(0,0) is an obvious solution, but there are more solutions.

note that

then

x+y=(x+y)(x-y)

so

0=(x+y)(x-y)-(x+y)

0=(x+y)(x-y-1)

so either x+y=0 or x-y-1=0

...
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Re: Determine 2 numbers w/ sum, products, diff. of sqrs equa

Postby Luke53 on Sun May 15, 2011 8:07 am

OK!

Tried out some more:

x y = (x + y) (x - y) and x y = x + y (given)
Subst. this in the first eqn. gives:
x y = x y (x - y) so x - y = 1 and x = y + 1 (same as : x - y - 1 = 0, see earlier by Martingale).
Subst. this in: x y = x + y (with x = y + 1) gives the quadratic eqn.: y² - y - 1 = 0 with solutions y = -0.618 or y = 1.618
Knowing that x = y + 1 gives:
x = .382 and y = - 0.618
x = 2.618 and y = 1.618
Seems that both results are solutions of the problem when worked out.
So thanks a lot for helping me out!
Great stuff!!!
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