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anonymous

  • 5 years ago

the number of real solutins of the equation -2-x^2+x=2^x

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  1. anonymous
    • 5 years ago
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    *solutions

  2. anonymous
    • 5 years ago
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    x=2

  3. anonymous
    • 5 years ago
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    Use the discriminant "\[b^2-4ac\]" if this is less than zero then there are no real roots, if greater than zero then there are 2 real roots , if x=0 , then there 1 real root. where b= the numer of x terms, in this case that is -1, a= the number of x^2 terms, in this case 3, and c= the number constants in this case 2.

  4. anonymous
    • 5 years ago
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    sorry i miss read the question, don't listen to me.

  5. anonymous
    • 5 years ago
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    the answer is that there is no real solution. i dont know how to figure that out

  6. anonymous
    • 5 years ago
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    so whats the answer?

  7. anonymous
    • 5 years ago
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    no real solution

  8. anonymous
    • 5 years ago
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    how about x=2 isn't it correct?

  9. anonymous
    • 5 years ago
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    if x=2 it gives the equality of that equation -2-2^2+2=2^2 or -2^2=2^2 or 4=4

  10. anonymous
    • 5 years ago
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    no -2^2=2^2 => -4=4 which is absurd

  11. anonymous
    • 5 years ago
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    what is the rule in multiplying two (-)number ?

  12. anonymous
    • 5 years ago
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    its not (-2)^2 its -(2^2)

  13. anonymous
    • 5 years ago
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    so that is not an exponent?

  14. anonymous
    • 5 years ago
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    what are you talking about?

  15. anonymous
    • 5 years ago
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    ok shall we say 3^2 so 2 is the exponent of 3.

  16. anonymous
    • 5 years ago
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    did this question ever get answered?

  17. anonymous
    • 5 years ago
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    It does not appear that it did... the number of real solutins of the equation \[-2-x^2+x=2^x\]

  18. anonymous
    • 5 years ago
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    how did you come up that equation?

  19. anonymous
    • 5 years ago
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    I copy and pasted from the original...

  20. anonymous
    • 5 years ago
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    so whats your final answer?

  21. anonymous
    • 5 years ago
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    I agree with the given, "no real solution"

  22. anonymous
    • 5 years ago
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    ok . am,what is real solution all about?

  23. anonymous
    • 5 years ago
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    my argument is that on the L.H.S. you have a polynomial degree 2 and on the R.H.S. you have the exponential expression \[2^x\] \[2^x\] is always positive and increasing so now we look at the L.H.S. more closely since it is a polynomial of degree 2, the highest degree term i.e. \[-x^2\] will dominate and thus the L.H.S. will be negative for "large" values of "x" now we observe that the quantity \[-x^2+x\] will only be positive when \[x\in(0,1)\] and moreover will never be greater than 2 Therefore the L.H.S. is always negative but the R.H.S' is always positve. Hence no real number solution i.e. if a solution does in fact exist to this equation it is not a real number value...

  24. watchmath
    • 5 years ago
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    Rewrite the equation as \(x^2-x+2+2^x=0\) Notice that \(x^2-x+2+2^x=(x-\frac{1}{2})^2+\frac{7}{4}+2^x >0\) So obviously the equation has no solution.

  25. anonymous
    • 5 years ago
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    Clearly.

  26. anonymous
    • 5 years ago
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    what exactly does it mean when it is said the an expression has a real solution. i mean whit is this "solution" all about?

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