Proof that a non-negative quadratic equation has at most one solution without using calculus

Question

I'm working through Lax and Terrell's 'Calculus with Applications'. One of the first exercises is to work through the proof of the Cauchy-Schwarz inequality. I'm all good for my proof except for the first part: i.e., how do I prove that a non-negative quadratic equation has at most one solution?

I can 'explain' this geometrically, i.e., a non-negative quadratic equation can only have a solution where the vertex touches the X axis. However, this doesn't feel very 'rigorous'.

Is there a better, algebraic, proof I could use here?

Thank you!

Andrew

Answer 1

And a THIRD option is to appeal to DeCartes Rule of Signs

over various case-by-case scenarios...

First off,

Q Must be positive, for otherwise f(x) will

be negative for large values of x...

So then there are 4 scenarios:

I) Q and R are both positive

In that case, the function is always positive,

so there are no solutions...

II) Q and R are both negative

In this case, there is ONE sign change, so

Decartes rule of signs says there is at most

one solution..

III-IV) Q and R are opposite signs...

If Q is positive and R is negative, then

there is only one sign change, which as mentioned,

guarantees only one solution, AT MOST

If Q is negative and R is positive...

Then there can be two solutions or none at all...

let P=R=1 and Q=-10...

then x^2-10x+1 is negative for values of x...

ex. x=1,2,3...

You would have to use an argument similar to

Daniel's to prove by contradiction this cannot happen

Answer 2

I think it depends a lot on what you allow yourself to assume.

First you need to assume a non-trivial quadratic equation.

For example, x² = x² has many solutions.

So let's rephrase the statement as

"A non-negative non-trivial quadratic polynomial has at most one real root".

Do you allow yourself to use the fact that such a polynomial is convex?

If so, then the existence of two distinct roots would imply that

all points between them lie below the x-axis,

contradicting non-negativity.

Those are my 2 cents.