Verification of correct solution to quadratic problem

[Mr Davis 97]

The problem statement: Show that if $r_1$ and $r_2$ are the distinct real roots of $x^2 + px + 8 = 0$, then $r_1 + r_2 > 4 \sqrt{2}$.

We start by noting that $r_1 r_2 = 8$. Using this relation, we'll find the minimum value of $r_1 + r_2$. To minimize $r_1 + r_2$, we need to minimize each term, thus we need to find the smallest values of both $r_1$ and $r_2$ such that $r_1 r_2 = 8$. The obvious answer is that $r_1 + r_2$ is minimized when $r_1, r_2 = 2 \sqrt{2}$, so that $r_1 + r_2 = 2 \sqrt{2} + 2 \sqrt{2} = 4 \sqrt{2}$ is the minimum value of the sum. However, we are under the condition that the roots are distinct, thus any other combination of roots would have to be greater than the minimum, showing that $r_1 + r_2 > 4 \sqrt{2}$.

Does this correctly show what the problem wants? Is there a flaw in the argument, or a better way to make the argument?

 

[Charles Link]

Looks good except a little calculus would be in order. And i think it assumes p<0. Anyway let $S=r_1+r_2=r_1+8/r_1$. Let $dS/dr_1=0$ ==>> $1-8/r_1^2=0$ so that $r_1^2=8$ and $r_1=r_2=2 \sqrt{2}$. (To show its a minimum $d^2 S/dr_1^2=16/r_1^3>0$ at $r_1=2 \sqrt{2})$. Otherwise your solution looks good. I'm not sure how much formal calculus you may have, etc. , but in any case your solution is quite good. (Note: Assuming it factors $(x-r_1)(x-r_2)=0$ with $r_1$ and $r_2>0$, this means p<0.)

 

[fresh_42]

I don't see why the minimum of $r_1+r_2$ has to be at a point where $r_1=r_2$. Why can't there be a solution (minimum) at a point $r_1-\epsilon$ and $r_2+\epsilon$?

(Charles beat me and has answered the question.)

 

[Mr Davis 97]

Well, it seemed intuitive to me that that value would be the minimum. Is it possible to show that $4 \sqrt{2}$ is the minimum without calculus (which I know a good bit of)? How could I make this proof without calculus in that case?

 

[Charles Link]

Well, it seemed intuitive to me that that value would be the minimum. Is it possible to show that $4 \sqrt{2}$ is the minimum without calculus (which I know a good bit of)? How could I make this proof without calculus in that case?

The function which you are finding the minimum, $S=r_1+8/r_1$ is non-trivial. If it were a quadratic, you could find the axis of symmetry. I do expect there is probably a way without calculus, but I don't know that there is an obvious one.

 

[fresh_42]

The function which you are finding the minimum, $S=r_1+8/r_1$ is non-trivial. If it were a quadratic, you could find the axis of symmetry. I do expect there is probably a way without calculus, but I don't know that there is an obvious one.

This is exactly my difficulty here. I'm sure there is an elegant argument anywhere to achieve it. But when your used to your toolbox, it's somehow like working with handcuffs.

 

[Mr Davis 97]

Here is my shot at a non-calculus approach:

Since we know that we have two distinct real roots, the discriminant will be greater than 0. Thus $\sqrt {p^2 - 32} > 0$, which means that $p > 4 \sqrt{2}$ or $p < -4 \sqrt{2}$. Thus, since $r_1 + r_2 = -p$, we have that $r_1 + r_2 > 4 \sqrt{2}$.

How is that?

 

[Charles Link]

@Mr Davis 97 Impressive that you came up with it so quickly ! :-) :-)

 

[fresh_42]

Here is my shot at a non-calculus approach:

Since we know that we have two distinct real roots, the discriminant will be greater than 0. Thus $\sqrt {p^2 - 32} > 0$, which means that $p > 4 \sqrt{2}$ and $p < -4 \sqrt{2}$. Thus, since $r_1 + r_2 = -p$, we have that $r_1 + r_2 > 4 \sqrt{2}$.

How is that?

Fine. I only think this solution is so close to simply calculating the roots, that I wonder why you shouldn't do this in the first place.
And again beaten. But I agree on Charles' compliment. Well done.

 

[Charles Link]

Here is my shot at a non-calculus approach:

Since we know that we have two distinct real roots, the discriminant will be greater than 0. Thus $\sqrt {p^2 - 32} > 0$, which means that $p > 4 \sqrt{2}$ or $p < -4 \sqrt{2}$. Thus, since $r_1 + r_2 = -p$, we have that $r_1 + r_2 > 4 \sqrt{2}$.

How is that?

It looks like the problem should read, prove $|r_1+r_2| >= 4 \sqrt{2}$ with absolute value signs.

 

[mfb]

It is a standard result that, for positive x,y,c, the sum of x and y for a given x*y=c is minimal if x=y. But the solution with the determinant is nicer.

Just don't forget the negative roots for positive p.

 

[Mr Davis 97]

Thanks for the help. It always nice to see how two different proofs relate to the same problem.

 

[symbolipoint]

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