Question

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

Answer 1

`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