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Quadratic with minimum values

  1. Oct 29, 2013 #1
    1. The problem statement, all variables and given/known data

    4. If x1, x2 are two real number roots of real number coefficient quadratic equation:
    $$x^2 -2mx + m + 2 =0$$
    Solve the following questions:
    (1) What are the values of m so that x1=x2?
    (2) What are the value(s) so that $$(x1)^2 + (x2)^2 $$
    has minimum value? What is the minimum?

    2. Relevant equations

    3. The attempt at a solution
    I substituted x1 into the quadratic for x and then substituted x2 into the quadratic for x and set them = to each other. I then solved for m and got
    m= (x1 + x2)/2

    I'm not sure what the 2nd question means. Can anyone help?
  2. jcsd
  3. Oct 29, 2013 #2


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    When x1 = x2, m is going to be some specific number, like 3.5, or [itex] \sqrt{2}[/itex] or something. Can you tell us what that number is and show us how you calculated it (or if not, show us how you got what you got).
  4. Oct 29, 2013 #3
    For a quadratic equation of the form ax2+bx+c=0, did they teach you what the sum of the roots is equal to in terms of a, b, and c? Did they teach you what the product of the roots is equal to?

    For part 1, can you use the quadratic formula to express the roots of the equation in terms of m?
  5. Oct 29, 2013 #4
    substitute x1 into the equation to get
    x1^2 -2mx1 +m+2
    Substitute x2 to get
    x2^2 -2mx2 +m+2
    set them = to each other and solve for m
    x1^2 -2mx1 + m + 2 = x2^2 -2mx2 +m + 2
    x1^2 - 2mx1 = x2^2 -2mx2
    x1^2 -x2^2 = m2(x1 - x2)
    then factor and divide to get
    m=(x1 + x2)/2
  6. Oct 29, 2013 #5
    Are you familiar with the formula:
    In terms of m, what are a, b, and c in your equation. Using the above equation, what value of m makes the two roots equal to one another?
  7. Oct 29, 2013 #6
    OK let me try that. add the roots and multiply by a to get b (keep the sign of b)and multiply the roots and multiply by a to get c?
  8. Oct 29, 2013 #7
    so a=1, b=-2m and c=m+2
    using the formula you get
    m±√m^2 - m + 2

  9. Oct 29, 2013 #8


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    There are brackets missing. The general idea is good, and a,b,c are right.
    After you fixed the brackets: Which value of m leads to the same value for x1 and x2?
  10. Oct 30, 2013 #9
    You made a mistake in algebra. That should be a minus 2 under the square root.
  11. Oct 30, 2013 #10
    Why you ask the OP to calculate the roots? The given questions doesn't require that.

    For part 1, it can be easily solved by investigating the discriminant of the given quadratic.

    For part 2, rewrite ##(x_1)^2+(x_2)^2## as ##(x_1+x_2)^2-2x_1x_2##. Its quite easy to solve from here.

    I am sorry if I said something wrong.
  12. Oct 30, 2013 #11
    Thanks Panav-Arora. Both these suggestions are what I originally had in mind. But the OP seemed to be struggling too much, so I thought I would try to make it easier on him. I thought that having him examine the solution using the quadratic formula would lead him to realizing that the discriminant should be set equal to zero.

    Part 2 is obviously a question in which he should have been thinking in terms of the sum of the roots, -b/a, and the product of the roots c/a. This must be what is being covered in his course right now. But he seemed to have no idea about this either. The equation you wrote above is exactly what I was thinking of.

  13. Oct 30, 2013 #12

    Ray Vickson

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    Since you have already done the question, I can now come in with a simpler solution.
    (1) the quadratic ##q(x) = 0## has a repeated root only if it is of the form ##q(x) = (x-r)^2##, or ##q(x) = x^2 - 2rx + r^2##. So in your case you need ##m + 2 = m^2##.
    (2) If ##x_1, x_2## are roots of ##x^2 = 2mx - m-2## then ##x_1^2 + x_2^2 = 2m(x_1 + x_2) - 2m - 4##, and ##x_1 + x_2 = 2m##. Thus, ##x_1^2 + x_2^2 = 4m^2 - 2m - 4.## This has the form y(y-1)-4, where y = 2m. Graphically, we see that the minimum of y(y-1) occurs at y = 1/2, giving m = 1/4. The minimum value of ##x_1^2 + x_2^2## is negative, meaning that x_1 and x_2 are complex.
    Last edited: Oct 30, 2013
  14. Oct 30, 2013 #13
    Yes I did make an error. Thank you.
  15. Oct 30, 2013 #14
    Ok after I fixed my error and set it = to 0 I get m= 2 or -1
  16. Oct 30, 2013 #15


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    Looks good.
  17. Nov 12, 2013 #16
    I took the derivative of 4m^2 - 2m -4 to get m=1/4. I showed this to my instructor and this is the reply I got.

    Good observation!

    then, you just consider m>=2 or m<=-1, and find the minimum value of function y=4m^2-2m-4. It is solvable

    I'm not sure what he means by this. I looked at graphs with m=3,4,5, -1,-2 and the minimum keeps getting lower. Any advice?
  18. Nov 12, 2013 #17


    Staff: Mentor

    Since you posted this problem in the Precalculus section, I am assuming that you are taking a class that comes before calculus. Using a technique from a later class is usually not appropriate, and in this case, not necessary.

    The graph of your function is a parabola that opens up, so its lowest point is at its vertex. You should be able to complete the square to write your equation as y = 4(x - h)2 + k. In this form, you can see by inspection that the vertex (low point) is at (h, k).
    See above.
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