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

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

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

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

Posts: 54
Joined: Sun Mar 13, 2011 9:46 am

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

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"?

maggiemagnet

Posts: 294
Joined: Mon Dec 08, 2008 12:32 am

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

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.
Luke53

Posts: 54
Joined: Sun Mar 13, 2011 9:46 am

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

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.

maggiemagnet

Posts: 294
Joined: Mon Dec 08, 2008 12:32 am

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

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

note that $x+y=x^2-y^2$

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

...

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

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

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.
Luke53

Posts: 54
Joined: Sun Mar 13, 2011 9:46 am