anonymous
  • anonymous
the number of real solutins of the equation -2-x^2+x=2^x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
*solutions
anonymous
  • anonymous
x=2
anonymous
  • anonymous
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.

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anonymous
  • anonymous
sorry i miss read the question, don't listen to me.
anonymous
  • anonymous
the answer is that there is no real solution. i dont know how to figure that out
anonymous
  • anonymous
so whats the answer?
anonymous
  • anonymous
no real solution
anonymous
  • anonymous
how about x=2 isn't it correct?
anonymous
  • anonymous
if x=2 it gives the equality of that equation -2-2^2+2=2^2 or -2^2=2^2 or 4=4
anonymous
  • anonymous
no -2^2=2^2 => -4=4 which is absurd
anonymous
  • anonymous
what is the rule in multiplying two (-)number ?
anonymous
  • anonymous
its not (-2)^2 its -(2^2)
anonymous
  • anonymous
so that is not an exponent?
anonymous
  • anonymous
what are you talking about?
anonymous
  • anonymous
ok shall we say 3^2 so 2 is the exponent of 3.
anonymous
  • anonymous
did this question ever get answered?
anonymous
  • anonymous
It does not appear that it did... the number of real solutins of the equation \[-2-x^2+x=2^x\]
anonymous
  • anonymous
how did you come up that equation?
anonymous
  • anonymous
I copy and pasted from the original...
anonymous
  • anonymous
so whats your final answer?
anonymous
  • anonymous
I agree with the given, "no real solution"
anonymous
  • anonymous
ok . am,what is real solution all about?
anonymous
  • anonymous
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...
watchmath
  • watchmath
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.
anonymous
  • anonymous
Clearly.
anonymous
  • anonymous
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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