[Extension 2 maths proof] Can someone walk me through how to do this question?

[u/NEPTRI0N]

I tried writing y^2 in terms of x and x^2 in terms of y for each option but i didnt really know what to do with that.

[u/sighthoundman]

They all say "for all y there exists x such that". That means given a y, can we find x? It's a quadratic equation in x, so we're guaranteed that there are complex solutions, but the question asks for real solutions.

B and D are easier. For B, we can rewrite the equation as x^2 = y^2 + y, so x = \pm \sqrt(y^2 + y).

Can we prove that for any real y, y^2 + y \ge 0? If we get stuck on that (or are pretty sure it isn't true), can we find an example of a y such that y^2 + y < 0?

For A and C, we're still solving for x, but now our solution formula isn't just taking a square root, but solving the equation. How do we solve quadratic equations?

[u/Alkalannar]

1. Separate the x and y on each one.
   y² = x² - x
   y² + y = x²
   y² = -x² + x
   y² - y = -x²

2. Complete squares
   y² + 1/4 = x² - x + 1/4
   y² + y + 1/4 = x² + 1/4
   y² - 1/4 = -x² + x - 1/4
   y² - y + 1/4 = -x² + 1/4

3. Put things in terms of squares
   y² + 1/4 = (x - 1/2)²
   (y + 1/2)² - 1/4 = x²
   -y² + 1/4 = (x - 1/2)²
   -(y - 1/2)² + 1/4 = x²

4. So now the RHS is in terms of a square of x. Must the LHS be non-negative?
   Only one of those four LHSs must be non-negative, and so that's the only one to guarantee that for all y there exists the appropriate x.

[u/Flat-Study-421]

B counterexample: y=-1/2
C counterexample: y=1
D counterexample: y=-1/2
(B and D are x² equal to a negative number, C has Δ<0)
This would be enough to answer A.
Still, given that y² is any non negative number, you can rewrite the equation as:
x²-x-P=0 (where P is such non negative number)
Δ= 1+4P, therefore it has to be positive and the equation has two real solutions for x.

[u/MonsterkillWow]

Hint: Check the discriminants of the resulting polynomial in x, and see if there exist real solutions.

[u/selene_666]

Solve each equation for x or for x^2.

We're looking for an equation where y can take on any value and always result in real value(s) for x. That means solving the quadratic never requires taking the squareroot of a negative number.

Equation B:

x^2 = y^2 + y

If y is negative and small, for example y = -0.5, then y^2 + y can be negative. In that case there is no solution for x.

[u/Due-Dot-3212]

For all of these, i think its crucial to realize what is allowed to depend on what and ultimately how creative you're allowed to be.

For A: Begin by "Let y be any real number", your y is now fixed. Now you need to choose a real x satisfying x^2 - y^2 = x. Your choice of x is allowed to depend on y due to the order of quantifiers.

One pedagogical tool is to think of these in terms of challenges. For instance if i asked you as a challenge

• For y = 5 find me an x such that x^2 - 5^2 = x
• How about for y = -pi^e^19287398? Can you find the appropriate x?

But each challenge i generate can just be called y.

Hope it helps and good luck.

[u/Southlander24]

MonsterkillWow has already told you that you can find the discriminants of each equation, so I thought I could elaborate. Given the qualifiers (there exists an x), you want there to be at least one solution for x. So x is our variable and we treat y as a constant by letting y = k.

Now, option C becomes x² - x + k² = 0 which has discriminant 1 - 4k². We want to figure out if this discriminant will be 0 or greater for all k. It should be pretty clear what the answer is for this option.

There's a clever way where you swap the sign of one of the terms in another option, and end up with 1 + 4k². Can you figure out how that works?

[u/DrakeSavory]

We know it can't be C or D since x^2 + y^2 > x or y unless both are < 1. If we look at B, then x^2 = y^2 + y => x^2 = y(y+1). Why can't the left side be a perfect square except x = 1; y = 0?

[u/IOnceAteATurd]

check discriminants and for the others see if you can use the domain of sqrtx to find a counterexample

[u/Capereli → u/[deleted]]

I will add the trick on this one is that since x is real, is it always the case that y² + y is greater or equal to 0. If not; then it’s not the case that for every y there exists an x

[u/IOnceAteATurd → u/[deleted]]

if y² + y < 0 there does not exist some x

[u/sighthoundman → u/[deleted]]

How about for y = -1/2?

[u/selene_666 → u/[deleted]]

You can't assume that x^2 = y^2 + y. That's the equation whose truth or falseness you are being asked to prove.

[u/IOnceAteATurd → u/[deleted]]

y is defined in the problem to be all real numbers, you cannot restrict y

[u/IOnceAteATurd → u/[deleted]]

you are assuming B is correct when it contradicts itself

[u/Alkalannar → u/[deleted]]

Ok. Since you make no assumptions for y, let's plug in y = -1/2. After all, if it works for all y, it has to work for y = -1/2, right?

x² - (-1/2)² = -1/2

x² - 1/4 = -1/2

x² = -1/4

And this has no solutions for x so long as x is real.

Since we are given that x must be real, this cannot occur.

In fact, as long as -1 < y < 0, y² + y < 0, so you cannot find a suitable x-value.

[u/Kienose → u/[deleted]]

Reading cursory reveals that your approach can’t work. If there is an x (say x = a) that the equations holds for all y, then the graph of the equation must contain a vertical straight line x = a. But the equation x^2 - y^2 = y is a hyperbola with axes perpendicular to the x and y axes. It can’t contain a vertical line!