Find the values of b for which the roots are positive

[Mr Davis 97]

Homework Statement

Given that $f(x) = x^2 - bx + 1$, find the values of b for which at least one of the roots are positive

Homework Equations

The Attempt at a Solution

So first I used the quadratic equation to find the roots: $\displaystyle x = \frac{b \pm \sqrt{b^2 - 4}}{2}$. Now, given these two roots, we need to find the intervals on which at least one is positive, so we need to find the union of the solution sets to $\displaystyle \frac{b + \sqrt{b^2 - 4}}{2} \ge 0$ and $\displaystyle \frac{b - \sqrt{b^2 - 4}}{2} \ge 0$. Looking at the first one, we end up with $b + \sqrt{b^2 - 4} \ge 0$. However, I am not quite sure how to solve this, which is where I get stuck.

 

[Mark44]

Homework Statement

Given that $f(x) = x^2 - bx + 1$, find the values of b for which at least one of the roots are positive

Homework Equations

The Attempt at a Solution

So first I used the quadratic equation to find the roots: $\displaystyle x = \frac{b \pm \sqrt{b^2 - 4}}{2}$. Now, given these two roots, we need to find the intervals on which at least one is positive, so we need to find the union of the solution sets to $\displaystyle \frac{b + \sqrt{b^2 - 4}}{2} \ge 0$ and $\displaystyle \frac{b - \sqrt{b^2 - 4}}{2} \ge 0$. Looking at the first one, we end up with $b + \sqrt{b^2 - 4} \ge 0$. However, I am not quite sure how to solve this, which is where I get stuck.

If $b > \sqrt{b^2 - 4}$ then both roots will be positive. What happens if $b = \sqrt{b^2 - 4}$ or if $b < \sqrt{b^2 - 4}$?

 

[Comeback City]

My first thought is Descartes Rule of Signs (I think it would work in some cases but not all). But I see another way. What type of value for "b" would always guarantee an [ x - (some positive root) ] ?

 

[Mr Davis 97]

If $b > \sqrt{b^2 - 4}$ then both roots will be positive. What happens if $b = \sqrt{b^2 - 4}$ or if $b < \sqrt{b^2 - 4}$?

Well I know that if $b > \sqrt{b^2 - 4}$ both roots will be positive, but I don't know how to solve for the solution set of b for which this is true. Also, if $b = \sqrt{b^2 - 4}$, then $0=-4$, which is false. If $b - \sqrt{b^2 - 4} < 0$, then at least one of the roots is negative

 

[Mark44]

Well I know that if $b > \sqrt{b^2 - 4}$ both roots will be positive, but I don't know how to solve for the solution set of b for which this is true.

Just solve the inequality.
$b > \sqrt{b^2 - 4} \Rightarrow b^2 > b^2 - 4$.
For what set of numbers b is this true?

Also, if $b = \sqrt{b^2 - 4}$, then $0=-4$, which is false. If $b - \sqrt{b^2 - 4} < 0$, then at least one of the roots is negative

For the equation, since 0 is not equal to -4, then the equation is never true. The last inequality is equivalent to $b < \sqrt{b^2 - 4}$. What do you get if you square both sides?

 

[Dick]

Well I know that if $b > \sqrt{b^2 - 4}$ both roots will be positive, but I don't know how to solve for the solution set of b for which this is true. Also, if $b = \sqrt{b^2 - 4}$, then $0=-4$, which is false. If $b - \sqrt{b^2 - 4} < 0$, then at least one of the roots is negative

Do a little more thinking and a little less solving. First of all if one root is positive you need that you have real roots. What's the condition for that? Start from there.

 

[Mr Davis 97]

Just solve the inequality.
$b > \sqrt{b^2 - 4} \Rightarrow b^2 > b^2 - 4$.
For what set of numbers b is this true?

For the equation, since 0 is not equal to -4, then the equation is never true. The last inequality is equivalent to $b < \sqrt{b^2 - 4}$. What do you get if you square both sides?

For $b > \sqrt{b^2 - 4}$, if we square both sides we get $b^2 > b^2 - 4$, so $0 > -4$, which is always true.This would seem to imply that the inequality is always true on the interval $(- \infty, -2] \cup [2, \infty)$. However, in actuality it is only true on the interval $[2, \infty)$. Is $(- \infty, -2]$ an "extraneous interval," just as we get extraneous solutions to radical equations? What's going on?

 

[Dick]

For $b > \sqrt{b^2 - 4}$, if we square both sides we get $b^2 > b^2 - 4$, so $0 > -4$, which is always true.This would seem to imply that the inequality is always true on the interval $(- \infty, -2] \cup [2, \infty)$. However, in actuality it is only true on the interval $[2, \infty)$. Is $(- \infty, -2]$ an "extraneous interval," just as we get extraneous solutions to radical equations? What's going on?

On $(- \infty, -2]$ your premise that $b > \sqrt{b^2 - 4}$ is false. So sure, it's extraneous But both those intervals are where you get real roots. Think about the signs of the roots on each of those intervals.

 

[Mr Davis 97]

On $(- \infty, -2]$ your premise that $b > \sqrt{b^2 - 4}$ is false. So sure, it's extraneous But both those intervals are where you get real roots. Think about the signs of the roots on each of those intervals.

If we're on the interval $(- \infty, -2]$, then both roots will be negative. If we're on the interval $[2, \infty)$, both roots are positive

 

[Dick]

If we're on the interval $(- \infty, -2]$, then both roots will be negative. If we're on the interval $[2, \infty)$, both roots are positive

Right. So problem solved, yes?

 

[pasmith]

Simple observation: If |b| \geq 2 then \sqrt{b^2 - 4} &lt; \sqrt{b^2} = |b|. Hence b - |b| &lt; b - \sqrt{b^2 - 4} <br /> \leq b + \sqrt{b^2 - 4} &lt; b + |b|.
If b \geq 2 then b - |b| = 0 and both roots are positive. If b \leq -2 then b + |b| = 0 and both roots are negative.

 

[epenguin]

Your first thought was right – you need Descartes's rule. That tells you that if b is positive any real roots any real roots are positive, while if b is negative any real roots are negative. After which the question is are there any real roots? As Dick said you needed more thinking less solving.

If you just think of what f(0), f(∞) and f(-∞) are that will also tell you you can't have one positive and one negative real root for this equation.