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How do you solve eq. that have both exponents and polynomials

  1. May 28, 2013 #1
    The question was: How many real number solutions are there for 2^x=-x^2-2x. I tired for an hour to isolate x but i couldn't do it. Then i used wolfram alpha and it gave me two solutions and graph. I realized that question was, how many not what are the solutions, and i could do that by graphing. But i still cant figure out algorithm for solving this type of equation.
     
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  3. May 28, 2013 #2

    mfb

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    There is no nice, analytic way to solve the equation.
     
  4. May 28, 2013 #3
    Intermediate value theorem is very powerful, you can use this to figure out the number of zeroes.
     
  5. May 28, 2013 #4

    Office_Shredder

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    If you want to be really hardcore about this you can do this without ever knowing how to graph any of these functions, just by knowing they are suitably differentiable. For example

    We want to find [itex] 2^x + x^2+ 2x = 0[/itex]

    Between every 0 of this function, the function has to change from increasing to decreasing. So the derivative needs to have a zero between each function. Taking the derivative
    [itex] \ln(2) 2^x + 2x + 2 = 0 [/itex]

    OK, between each of the zeros here the derivative has to be zero. So we take the second derivative
    [itex] \ln(2)^2 2^x + 2 = 0[/itex]
    This has no solutions because the left hand side is always positive. So this means that
    [itex] \ln(2) 2^x + 2x + 2 = 0[/itex]
    has at most one solution. We can observe that it has at least one solution because plugging in x=-infinity and x=infinity it changes from positive to negative. This means that
    tex] 2^x + x^2+ 2x = 0 [/tex]
    Can have at most two solutions. If x is incredibly large in magnitude (negative or positive), then the function is positive, so there are two zeros if we can find an x making the left hand side negative, and no zeros if we can never make the left hand side negative.

    How can we perform this search for a value of x making [itex] 2^x + x^2 + 2x[/itex] negative? Well, the 2x is positive, as is x2. Ff x is very far from 0 (negative), then 2x is negligible and the x2 is the dominating positive term. So we can find bounds on where any potential negative value come from by checking:
    [tex] x^2 + 2x > 0[/tex]
    Occurs whenever x < -2
    So we need to find a number between -2 and 0 making the function 0. As a first guess we try x = -1
    [tex] 2^{-1} + (-1)^2 - 2 = -1/2 < 0[/tex]
    And we've shown there are two solutions! Even better we now know that one of them is between -2 and -1, and the other is between -1 and 0
     
  6. May 29, 2013 #5
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