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Find the values of b for which the roots are positive

  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    Given that ##f(x) = x^2 - bx + 1##, find the values of b for which at least one of the roots are positive

    2. Relevant equations


    3. 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.
     
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  3. Mar 21, 2017 #2

    Mark44

    Staff: Mentor

    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}##?
     
  4. Mar 21, 2017 #3
    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) ] ?
     
  5. Mar 21, 2017 #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
     
  6. Mar 21, 2017 #5

    Mark44

    Staff: Mentor

    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?
     
  7. Mar 22, 2017 #6

    Dick

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    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.
     
  8. Mar 23, 2017 #7
    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?
     
  9. Mar 23, 2017 #8

    Dick

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    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.
     
  10. Mar 23, 2017 #9
    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
     
  11. Mar 23, 2017 #10

    Dick

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    Right. So problem solved, yes?
     
  12. Mar 23, 2017 #11

    pasmith

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    Simple observation: If [itex]|b| \geq 2[/itex] then [itex]\sqrt{b^2 - 4} < \sqrt{b^2} = |b|[/itex]. Hence [tex]b - |b| < b - \sqrt{b^2 - 4}
    \leq b + \sqrt{b^2 - 4} < b + |b|.[/tex]
    If [itex]b \geq 2[/itex] then [itex]b - |b| = 0[/itex] and both roots are positive. If [itex]b \leq -2[/itex] then [itex]b + |b| = 0[/itex] and both roots are negative.
     
  13. Mar 28, 2017 #12

    epenguin

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    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.
     
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