Proof that two simultaneous line equations do not intersect?

Question

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:

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?

Answer 1

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.

Answer 2

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.)

Answer 3

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.