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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ankur, This is a quadratic equation, \[x^2+2x2=0\]To determine the nature of the solutions to a quadratic, we use something called the 'discriminant', defined as\[\Delta = b^24ac\]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\[\Delta = 2^24(1)(2)=12>0\]which means your quadratic will have two real solutions.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0The function \[2^x2+x^2\] is continuous, negative at 0 and positive at both 2 and 2. Thus there are roots in those two intervals.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0I hope I have copied the equation correctly. The equation is \[2^{x}=2x ^{2}\] I set x to =2 I get the following\[2^{2}=22^{2}\]

radar
 5 years ago
Best ResponseYou've already chosen the best response.04=24 Duh I don't think so if I let x=0 then I have 1=21 hey 1=1 that is a solution! What about x=1, then 2 =22=0 No 1 is not a solution. If x=3 then 8=29 8=7 No What is the solution and is the term \[2^{x}\] makes question this problem unless is something do with logs.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Only that 0^2 = 0 and not 1...

radar
 5 years ago
Best ResponseYou've already chosen the best response.0I assume you were discussing when I let x = 0

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Yes indeed and you mixed up something. If you insert x=0 you get \[2^0 = 2  0^2 ⇔ 1 = 2  0\] which is false.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Ah yes nowhereman, so even 0 is not a solution!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, Ankur, I read your equation incorrectly.
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