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Homework Help: Find the value of k

  1. Dec 10, 2011 #1
    1. The problem statement, all variables and given/known data

    For the equation 4(times)sqrtx= 2x+k find a value such that k the equation has (a) one solution, (b) two solutions, (c) no solutions

    2. Relevant equations


    3. The attempt at a solution

    From original equation, I got to:



    x= (4x^2+4xk+k^2)/16

    simplified to: x= 1/4x^2 +1/4xk + 1/16k^2

    Put it into quadratic formula and got the discriminant as: sqrt((1/4k)^2 -(4)(1/4)(k^2/16))

    Dont know what to do next... Please help and show all work. Thank You
  2. jcsd
  3. Dec 10, 2011 #2


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    You don't have the equation, x= 1/4x^2 +1/4xk + 1/16k^2, in the correct form to use the quadratic formula. There's an x on the left hand side.

    More simply:
    Square the original equation, [itex]\displaystyle 4\sqrt{x}= 2x+k[/itex]

    giving [itex]\displaystyle 16x= 4x^2+4kx+k^2[/itex]

    [itex]\displaystyle 4x^2+(4k-16)x+k^2=0[/itex]​

    Use the discriminant, b2-4ac, to determine the value of k needed for 0, 1, or 2 real solutions.
  4. Dec 10, 2011 #3
    Oh, thanks, So for 1 solution the discriminant would have to be 0, and for 2 solutions ≤0 and what would it have to be for 0 solutions? just ≥0? becuase cant u still get 2 imaginary solutions?
  5. Dec 10, 2011 #4
    the final answer I got was: k=25/128 for one solution k>25/128 for no solutions and k<25/128 for 2 real solutions
  6. Dec 10, 2011 #5


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    No, not quite.
    If b2-4ac = 0, there is 1 real solution.
    If b2-4ac > 0, there are 2 real solutions.
    If b2-4ac < 0, there are no real solutions (but there are 2 complex solutions).
  7. Dec 10, 2011 #6
    oh yeah, sorry, I accidentally included the "equal to" but I fixed it in the final answer..
  8. Dec 10, 2011 #7


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    The "equal to" is not the only thing that's incorrect in the bolded part above.
  9. Dec 10, 2011 #8
    oh, it wasnt bolded in previous part, but i see, >0 is 2 solutions, <0 no solutions( complex) = 0 one solution
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