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anonymous

  • 5 years ago

Show that the equation 2^x= 2 − x^2 has at least two real roots??

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  1. anonymous
    • 5 years ago
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    Ankur, This is a quadratic equation, \[x^2+2x-2=0\]To determine the nature of the solutions to a quadratic, we use something called the 'discriminant', defined as\[\Delta = b^2-4ac\]where a,b, and c are the coefficients of the quadratic,\[ax^2+bx+c=0\]A quadratic has two real solutions if\[\Delta >0\] one real solution when\[\Delta = 0\]and no real solutions when\[\Delta <0\]In your case\[a=1,b=2,c=-2\]so\[\Delta = 2^2-4(1)(-2)=12>0\]which means your quadratic will have two real solutions.

  2. nowhereman
    • 5 years ago
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    The function \[2^x-2+x^2\] is continuous, negative at 0 and positive at both -2 and 2. Thus there are roots in those two intervals.

  3. radar
    • 5 years ago
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    I hope I have copied the equation correctly. The equation is \[2^{x}=2-x ^{2}\] I set x to =2 I get the following\[2^{2}=2-2^{2}\]

  4. radar
    • 5 years ago
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    4=2-4 Duh I don't think so if I let x=0 then I have 1=2-1 hey 1=1 that is a solution! What about x=1, then 2 =2-2=0 No 1 is not a solution. If x=3 then 8=2-9 8=-7 No What is the solution and is the term \[2^{x}\] makes question this problem unless is something do with logs.

  5. nowhereman
    • 5 years ago
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    Only that 0^2 = 0 and not 1...

  6. radar
    • 5 years ago
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    That was a \[2^{0}\]

  7. radar
    • 5 years ago
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    I assume you were discussing when I let x = 0

  8. nowhereman
    • 5 years ago
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    Yes indeed and you mixed up something. If you insert x=0 you get \[2^0 = 2 - 0^2 ⇔ 1 = 2 - 0\] which is false.

  9. radar
    • 5 years ago
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    Ah yes nowhereman, so even 0 is not a solution!

  10. anonymous
    • 5 years ago
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    Sorry, Ankur, I read your equation incorrectly.

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