Find the number of real root and the number of imaginary root in the following equation #e^x=x^2# =?

1 Answer
Jul 12, 2016

Answer:

#x = -0.703467#

Explanation:

#e^x = (e^{x/2})^2#

so

#e^x - x^2= (e^{x/2})^2-x^2 = (e^{x/2}+x) (e^{x/2}-x)=0#

Analyzing the solutions

1) #e^{x/2}+x = 0->e^{x/2}=-x#
but #e^{x/2} > 0, forall x in RR# so the solution must be a negative number. Solving iteratively we obtain
#x = -0.703467#

2) #e^{x/2}-x=0->e^{x/2}=x#
but #e^{x/2}>x, forall x in RR# so no intersection between #y = e^{x/2}# and #y = x# and consequently, no solution.

Concluding. The solution is #x = -0.703467#

Attached a figure showing:
#y = e^{x/2}# green
#y = x# red
#y = -x# blue
and the solution point in black

enter image source here