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Apologies if this isn't at the level of questions expected here!

I've got two simultaneous equations to solve.

(Equation 1): $ x y = 4 $

(Equation 2): $ x + y = 2 $

They produce the following curves:

enter image description here

Question: Whilst it's graphically obvious that they do not make contact, what is the algebraic indicator that these two lines do not meet? How do you prove that?

yellow-saint
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  • Nice graph. Early on, you may want Equation 2) to be $x+y=2$, or something equivalent. – André Nicolas Sep 20 '14 at 17:40
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    Whoops, that was infact the equation I meant to type @AndréNicolas - I got muddled figuring out what MathJaX was :) – yellow-saint Sep 20 '14 at 17:43
  • One algebraic indicator is the sign of the discriminant of the polynomial $x(2-x)-4$. Alternately, use $(x-y)^2=(x+y)^2-4xy$ to conclude that if $(x,y)$ is a meeting point then $(x-y)^2=4-16$. Impossible. (But I find the picture enough!) – André Nicolas Sep 20 '14 at 17:46

3 Answers3

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So the curves are

$$\frac4x\quad\text{and}\quad2-x$$

To prove that these curves do not meet, it simply means that

$$\frac4x=2-x$$

has no solutions, to solve this we multiply by $x$

$$4=2x-x^2$$

Then multiply with $-1$ and switch the left and right of the equality

$$x^2-2x=-4$$

Then I'm going to add one to each side

$$x^2-2x+1=-3$$

The left hand side can be rewritten as

$$(x-1)^2=-3$$

No number squared can be negative, so the equation has no solutions.

Alice Ryhl
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    I would just stop at $(x-1)^2 = -3$ and conclude from here as no square can be negative. – Winther Sep 20 '14 at 17:45
  • @Winther Good point – Alice Ryhl Sep 20 '14 at 17:47
  • Thanks very much for this answer. I've 'ticked' this one just because it's the one that's most at my level of understanding, but I'm sure the others are just as robust :) So the general rule is, try to express the standard quadratic form of the equation, and then go from there? – yellow-saint Sep 20 '14 at 17:58
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    @J.B. The general rule is set them equal and try solve it, if you get solutions, you found places they intersect, if you find no solutions, they dont intersect – Alice Ryhl Sep 20 '14 at 17:59
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Suppose that the curves meet: there are $x,y$ such that $xy=4$ and $x+y=2$. Then, $x$ and $y$ are real roots to the polynomial $$ P(z)=z^2-2z+4. $$ But $P(z)=(z-1)^2+3$ clearly has no real roots, so contradiction. (Ref: Vieta's formulas.)

Kim Jong Un
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If the two curves intersect (meet), they do so wherever they have a point in common. This will only happen when $y_1=y+2$, \iff $$\frac 4x = 2-x \iff 2x - x^2 = 4 \iff x^2 -2x + 4 = 0$$ for some real $x$.

Use the quadratic equation to show that there is no real solution to this equation Indeed, you need only check the discriminant of the quadratic $$\underbrace{4 - 16}_{b^2 - 4ac} = -12\lt 0 $$ to see that it is undefined in the reals, and hence no real $x$ that make this equation true.

I.e., the lines cannot intersect.

amWhy
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