1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to find a constant in this quadratic equation?

  1. Jun 16, 2017 #1
    1. The problem statement, all variables and given/known data
    2014_Mat_B.png

    2. Relevant equations
    for equation which has 2 different solutions, D >0


    3. The attempt at a solution
    (1)
    D > 0
    b^2 - 4ac > 0
    3 - 4root2.k > 0
    k < 3 / ( 4root 2 )
    k < ( 3 root 2 ) /8

    has solution of sin tetha and cos tetha
    sin 0 = 0, cos 0 = 1.
    when x = 0, and x = 1 --> satisfy the rule?

    when x = 0,
    k = 0...
    when x = 1,
    k = 1/(root 3 - root 2)

    can anyone help me to finish it?

    (2) (x^3 + a/x^2)^5 = -270
    something to the power 5 would be -270, means that something is negative
    the factor of 270 = 3 x 3 x 3 x 5 x 2
    can anyone help me to finish it?

    (3) f( g(x) ) = 3gx + 1 / (2gx + 1)
    x = 3gx + 1 / (2gx + 1)

    and finally I get
    2x^2 - 3x + 3 = -2px^2 + 3px + 3
    (x1 + x2) on left equation = (x1 + x2) on right ?

    -b/a = -b/a
    3/2 = -3p/-2p

    and x1.x2 on left equation = x1.x2 on right ?
    c/a = c/a
    3/2 = 3/-2p
    p = -1
    Is it right way?
     
  2. jcsd
  3. Jun 16, 2017 #2
    Hello! please use latex next time thanks!

    1) Let x1=sin(θ) and x2=cos(θ), then by substitution into the original equation you can find the value of k since you are given the interval of θ.

    2) Im not sure what is meant by the "constant term" but I will asume is [itex](x^3+ \frac{a}{x^2})^5=-270[/itex].
    Then by [itex]-(x^3+ \frac{a}{x^2})=(270)^5[/itex] and therefore by multipling by [itex]x^2[/itex] we have [itex](x^5+ a)=-x^2(270)^5[/itex] rearranging factors [itex]x^2(x^3+(270)^5)=-a[/itex] shows that a depends on a so there is not a definite answer but an infinite of them... give x a value and you get a value of a and viceversa.

    3) By substitution let x in f(x) be [itex]x=\frac{px+1}{2x-3}[/itex] and then impose f(x)=x. As a hint, in the interval x≠-1/2,3/2 you would probably end up having an expresion of the type 1/0...

    Hope to help.
     
  4. Jun 17, 2017 #3
    I still not find number 2.. can anyone help?
     
  5. Jun 17, 2017 #4

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    Expand (x3 +a/x2)5 using the binomial theorem and look at the constant term.
     
  6. Jun 17, 2017 #5
    using binomial...
    $$X^{15} - 5x^{12} \frac {a} {x^2} + 10x^{9} \frac {a^{2}} {x^{4}} - 10x^{6} \frac {a^{3}} {x^{6}} + 5x^{3} \frac {a^{4}} {x^{8}} - \frac {a^{5}} {x^{10}} = -270 $$

    now I don't know..
    can anyone help?
     
  7. Jun 17, 2017 #6

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    Double check your binomial expansion. It is wrong. Then simplify the terms. Combine the x factors in each term by combining the exponents.
     
  8. Jun 17, 2017 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Why are the signs alternating?

    Which term is the constant term ?
     
  9. Jun 17, 2017 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That term in the binomial expansion which is independent of x.
    No, only the constant term equals -270, not the whole function.
     
  10. Jun 17, 2017 #9
    I think it doesnt have constant term. The last term is $$\frac { a^5} {x^{10}}$$
     
  11. Jun 17, 2017 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The constant term need not be the last one.
     
  12. Jun 17, 2017 #11

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    Simplify each term by combining the exponents of x like xn/xm = xn-m. One term will end up with x0 = 1. That is the constant term. I shouldn't help more than that on a homework problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: How to find a constant in this quadratic equation?
Loading...