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

Jul 12, 2016

$x = - 0.703467$

#### Explanation:

${e}^{x} = {\left({e}^{\frac{x}{2}}\right)}^{2}$

so

${e}^{x} - {x}^{2} = {\left({e}^{\frac{x}{2}}\right)}^{2} - {x}^{2} = \left({e}^{\frac{x}{2}} + x\right) \left({e}^{\frac{x}{2}} - x\right) = 0$

Analyzing the solutions

1) ${e}^{\frac{x}{2}} + x = 0 \to {e}^{\frac{x}{2}} = - x$
but ${e}^{\frac{x}{2}} > 0 , \forall x \in \mathbb{R}$ so the solution must be a negative number. Solving iteratively we obtain
$x = - 0.703467$

2) ${e}^{\frac{x}{2}} - x = 0 \to {e}^{\frac{x}{2}} = x$
but ${e}^{\frac{x}{2}} > x , \forall x \in \mathbb{R}$ so no intersection between $y = {e}^{\frac{x}{2}}$ and $y = x$ and consequently, no solution.

Concluding. The solution is $x = - 0.703467$

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