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

Please help.

Luke.

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

Please help.

Luke.

Please help.

Luke.

- maggiemagnet
**Posts:**358**Joined:**Mon Dec 08, 2008 12:32 am-
**Contact:**

Try using variables: x+y = xy = xDetermine two numbers: their sum, their product and the difference of their squares are equal.

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

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.

Luke.

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.

Luke.

- maggiemagnet
**Posts:**358**Joined:**Mon Dec 08, 2008 12:32 am-
**Contact:**

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

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.

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.

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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(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

...

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

...

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!!!

Luke.

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!!!

Luke.