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For what values of k will the equation have no real roots?

  1. Apr 21, 2015 #1
    1. The problem statement, all variables and given/known data
    2x^2-3x+kx=-1/2

    1. k<1 or k>1
    2. 1<=k<=5
    3. k<=1 or k>=5
    1<k<5

    2. Relevant equations
    b^2-4ac
    a=2 b=3 c=k
    3. The attempt at a solution
    (3)^2-4(2)(k)
    =9-8k<0
    =9/8<k
    =1&1/8<k

    I get the answer above but don't know how it relates???
    Any insight would be appreciated.

    Thank you,
    Jaco
     
  2. jcsd
  3. Apr 21, 2015 #2

    MarcusAgrippa

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    Find the discriminant of your quadratic equation. The equation has no real roots if the discriminant is less than zero.
     
  4. Apr 21, 2015 #3

    jbriggs444

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    Look at the original equation again.

    What is the coefficient on x? What is the constant term?
     
  5. Apr 21, 2015 #4

    MarcusAgrippa

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    Are you sure about your value for b?
     
  6. Apr 21, 2015 #5
    Jbriggs,
    I have been looking at other threads and found a similar example and there:
    (-3)^2-4(2)(k+1/2)
    =-9-8k+4
    =-5-8k
    => k=5/8
    Thank you for pointing that out Marcus
     
  7. Apr 21, 2015 #6
    Is this correct?
     
  8. Apr 21, 2015 #7

    MarcusAgrippa

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    It doesn't look correct. What is the general form of a quadratic equation? Write your equation in a way that imitates that general form and identify correctly the values of a,b,c. Then write down the general definition of the discriminant, and substitute your values of a,b,c into it. The solve the inequality (discriminant) < 0 for k.
     
  9. Apr 21, 2015 #8

    SammyS

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    Please restate the entire problem as it was given to you.

    You may have a typo in the quadratic expression as you posted it. If it's correct as it is, then you still do not have the correct b or c .
     
  10. Apr 21, 2015 #9
    Sammy,
    For what values of k will the equation 2x^2 -3x + kx = -1/2 have no real roots?


    Possible answers:
    1. k<1 or k>1
    2. 1<=k<=5
    3. k<=1 or k>=5
    1<k<5
     
  11. Apr 21, 2015 #10

    jbriggs444

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    Everyone responding to this thread is attempting to point out that the b and c that you have harvested from that equation are wrong.
     
  12. Apr 21, 2015 #11
    ok, I have been considering this too as:
    (3x+kx)^2-4(2)(-1/2)
    (3x+kx)(3x+kx)+4
    9x^2+3kx^2+3kx^2+k^2x^2
    9x^2+6kx^2+k^2x^2
     
  13. Apr 21, 2015 #12

    jbriggs444

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    Re-read post #7 above. What is the standard form for a quadratic equation? Can you restate the original equation in that form?
     
  14. Apr 21, 2015 #13
    ax^2+bx+c=0
    2x^2+(3x+kx)+1/2=0
    2x^2+3x+kx=-1/2
    kx=-1/2-2x^2-3x
    k=(1/2-2x^2-3x)/x
     
  15. Apr 21, 2015 #14
    wow, i feel more confused...
     
  16. Apr 21, 2015 #15

    SammyS

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    It's good to this point.

    Take the expression in parentheses and factor out x.
     
  17. Apr 21, 2015 #16
    2x^2+(3x+kx)+1/2=0
    2x^2+x(3+k)+1/2=0
    2x^2+(3+k)(x+1/2)=0
    2x^2+3x+1&1/2+kx+1/2k=0
    kx+1/2k=-2x^2-3x-1&1/2
    3/2kx=-2x+3+1&1/2
    3kx=-4x+9
    k=(-4x+9)/x
     
    Last edited: Apr 21, 2015
  18. Apr 21, 2015 #17

    SammyS

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    Stop at this point !
    What is the coefficient of x ?
     
  19. Apr 21, 2015 #18
  20. Apr 21, 2015 #19

    jbriggs444

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    In the expression 2x2 + 3x + 4, what is the coefficient on the "x" term?
     
  21. Apr 21, 2015 #20
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