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

1. anonymous

*solutions

2. anonymous

x=2

3. 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.

4. anonymous

sorry i miss read the question, don't listen to me.

5. anonymous

the answer is that there is no real solution. i dont know how to figure that out

6. anonymous

7. anonymous

no real solution

8. anonymous

how about x=2 isn't it correct?

9. 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

10. anonymous

no -2^2=2^2 => -4=4 which is absurd

11. anonymous

what is the rule in multiplying two (-)number ?

12. anonymous

its not (-2)^2 its -(2^2)

13. anonymous

so that is not an exponent?

14. anonymous

15. anonymous

ok shall we say 3^2 so 2 is the exponent of 3.

16. anonymous

did this question ever get answered?

17. anonymous

It does not appear that it did... the number of real solutins of the equation $-2-x^2+x=2^x$

18. anonymous

how did you come up that equation?

19. anonymous

I copy and pasted from the original...

20. anonymous

21. anonymous

I agree with the given, "no real solution"

22. anonymous

ok . am,what is real solution all about?

23. 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...

24. 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.

25. anonymous

Clearly.

26. 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?