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anonymous
 5 years ago
the number of real solutins of the equation 2x^2+x=2^x
anonymous
 5 years ago
the number of real solutins of the equation 2x^2+x=2^x

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use the discriminant "\[b^24ac\]" 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i miss read the question, don't listen to me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is that there is no real solution. i dont know how to figure that out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how about x=2 isn't it correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if x=2 it gives the equality of that equation 22^2+2=2^2 or 2^2=2^2 or 4=4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no 2^2=2^2 => 4=4 which is absurd

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is the rule in multiplying two ()number ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its not (2)^2 its (2^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so that is not an exponent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what are you talking about?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok shall we say 3^2 so 2 is the exponent of 3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did this question ever get answered?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It does not appear that it did... the number of real solutins of the equation \[2x^2+x=2^x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you come up that equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I copy and pasted from the original...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so whats your final answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I agree with the given, "no real solution"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok . am,what is real solution all about?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my 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...

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Rewrite the equation as \(x^2x+2+2^x=0\) Notice that \(x^2x+2+2^x=(x\frac{1}{2})^2+\frac{7}{4}+2^x >0\) So obviously the equation has no solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what 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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