Solving for Real Values of x, y, & z with x+y=1 and xy-z^2=1

[Euler_Euclid]

find all real values such that:
$$x+y=1$$
and $$xy-z^2=1$$

This one was in one of the exams we had.

 

[HallsofIvy]

Nothing at all difficult about that, except that I presume you mean "find all 'triples' of real numbers, (x, y, z) that satisfy the equations". From x+ y= 1, y= 1- x. Putting that for y in the second equation, x(1- x)- z²= x- x²- z²= 1 which we can rewrite as the quadratic equation x²- x- (z²+ 1)= 0. Now we can solve that for x, as a function of z, using the quadratic formula:
$$x= \frac{-(-1)\pm\sqrt{(-1)^2- 4(1)(-(z^2+ 1)}}{2}= \frac{1\pm \sqrt{8z^2+ 9}}{2}$$
Now, since 8z²+ 9 is positive for all z, z can be any number, x can be calculated from that formula, and then y can be calculated from y= 1- x.

 

[gel]

$$x= \cdots= \frac{1\pm \sqrt{8z^2+ 9}}{2}$$

As x must be positive, that can't be right.

The problem is easy to see graphically. The set of values for (x,y) is just the intersection of the region xy>=1 with the line x + y = 1, which gives a closed line segment.

 

[disregardthat]

$$x^2+2xy+y^2=1$$
$$x^2+y^2+2(1+z^2)=1$$
$$x^2+y^2+2z^2=-1$$

or

We observe that both y and x must be positive.

$$(x+y)^2/4>=xy$$, so $$xy<=1/4$$. (AM-GM)
$$z^2=xy-1<=-3/4$$, so no solutions for z if x and y are real.

 

[disregardthat]

As x must be positive, that can't be right.

The problem is easy to see graphically. The set of values for (x,y) is just the intersection of the region xy>=1 with the line x + y = 1, which gives a closed line segment.

How on Earth can you deduce that?

 

[Hurkyl]

How on Earth can you deduce that?

z² is nonnegative.

 

[disregardthat]

Of course.
Intersection of the region xy>=1 and the line x+y=1 :

y>=1/x, and y=1-x Suppose 1-x > 1/x, then 1/x+x<1 which is impossible! (as x is positive)

 

[Hurkyl]

Right. I find it slightly quicker to visualize the result -- xy >= 1 fills in the top-right and bottom-left parts of the hyperbola xy=1... and x+y=1 passes right between the two branches.

 

[disregardthat]

Either way, the closed line segment doesn't exist, right?

 

[praharmitra]

The question has no solution... Consider the following argument...


x + y = 1
xy = 1 + z^2

Now if x > 1, then
y = 1 - x => y < 0

This gives x > 1, y < 0, Thus xy < 0. But since xy = 1 + z^2, xy must be positive.

Thus x <= 1 and y <= 1

Now if x < 0, then
y = 1 - x => y > 1

In this case x < 0 and y > 1. Therefore xy < 0.
Thus x >= 0 and y >= 0.

Taking in both arguments
0 <= x <= 1
and 0 <= y<= 1.

In this case xy <= 1. However since xy = z^2 + 1, xy must be greater than or equal to 1. The only possibility is xy = 1

Solve for

x+y = 1
xy = 1

Since no real value satisfies this equation, no solution exists

 

[praharmitra]

Nothing at all difficult about that, except that I presume you mean "find all 'triples' of real numbers, (x, y, z) that satisfy the equations". From x+ y= 1, y= 1- x. Putting that for y in the second equation, x(1- x)- z²= x- x²- z²= 1 which we can rewrite as the quadratic equation x²- x- (z²+ 1)= 0

Well, u made a mistake. x(1-x) - z² = 1 does not reduce to x^2 - x - (z^2 - 1) = 0. Infact, the equation is $$x^2 - x + z^2 + 1 = 0$$.

$$D = (-1)^2 - 4(1)(z^2 + 1) = 1 - 4 - 4z^2 = -3 - 4 z^2$$ which is negative for all values of z. Thus, no real solution exists

 

[arildno]

Thanks for clearing that up,prahamirta.
Another way of seeing this is by rewriting the correct quadratic as
$$(x-\frac{1}{2})^{2}+z^{2}=-\frac{3}{4}$$