How to solve for x? (2/9)x^2 + (4/3)x > 0.5^x

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The discussion centers on solving the inequality (2/9)x^2 + (4/3)x > 0.5^x, where participants express difficulty in finding an algebraic solution. It is noted that plotting the functions reveals they intersect around x=0.5, not at x=0, and numerical analysis suggests a solution near x=0.4. Participants agree that while numerical methods can yield results, an algebraic solution using standard functions is not feasible. The conversation concludes with a query about the existence of any unusual functions that could assist in solving the inequality.
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Hi all, for what I'm currently doing I need to find where that parabola is greater than the exponential equation.This may seem simple but I can't seem to do anything other than rephrase it without getting a solution. Any help appreciated. it seems as though the answer is where x>0 but I can't prove it algebraicly.
 
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Superposed_Cat said:
Hi all, for what I'm currently doing I need to find where that parabola is greater than the exponential equation.This may seem simple but I can't seem to do anything other than rephrase it without getting a solution. Any help appreciated. it seems as though the answer is where x>0 but I can't prove it algebraicly.
With over 300 posts, you should know by now not to put your equations in the title.

If you plot the functions on the two sides of the inequality you will see they cross near ##x=1/2##, not near ##x=0##. You can solve for it numerically.
 
Numerical analysis gives about 0.4, you are right. But is it undoable algebraicaly?
 
Superposed_Cat said:
Numerical analysis gives about 0.4, you are right. But is it undoable algebraicaly?

It is closer to .5 than .4. Yes, you can not solve it algebraically in terms of the usual functions.
 
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Thank you, like your signature by the way. "usual functions" is there an unusual one to do this job?
 

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