Solve: ${e^{2x}} - 5{e^x} + 6 = 0$.

Hint:The given problem requires us to solve an equation. The given equation can be reduced to a simple quadratic equation using a substitution. There are various methods that can be employed to solve a quadratic equation like completing the square method, using quadratic formulas and by splitting the middle term.

Complete step by step solution:
In the given question, we are required to solve the equation ${e^{2x}} - 5{e^x} + 6 = 0$ .

Consider the equation ${e^{2x}} - 5{e^x} + 6 = 0$.

Let ${e^x} = t - - - - - - (1)$

Then, ${e^{2x}} = {t^2}$

So, the given equation ${e^{2x}} - 5{e^x} + 6 = 0$becomes $t{}^2 - 5t + 6 = 0$ .

Now, considering equation $t{}^2 - 5t + 6 = 0$, the equation can be solved by various methods such as completing the square method, splitting the middle term and using the quadratic formula.

Solving the quadratic equation by splitting the middle term, we get,

\[ = {t^2} - \left( {3 + 2} \right)t + 6 = 0\]
$ = {t^2} - 3t - 2t + 6 = 0$
$ = \left( {{t^2} - 3t} \right) + \left( { - 2t + 6} \right) = 0$
$ = t\left( {t - 3} \right) - 2\left( {t - 3} \right) = 0$
$ = (t - 3)(t - 2) = 0$

Now, either $\left( {t - 3} \right) = 0$ or $\left( {t - 2} \right) = 0$
So, either $t = 3$ or $t = 2$

Substituting the value of t back, we get,

Either ${e^x} = 3$ or ${e^x} = 2$

Now we have two exponential equations to solve and find the values of x. We can solve the exponential equations by taking log both sides and simplifying the calculations.

So, either ${e^x} = 3$ or ${e^x} = 2$

$\log \left( {{e^x}} \right) = \log 3$ or $\log \left( {{e^x}} \right) = \log 2$

$x = \log 3$ or $x = \log 2$

So, the roots of the equation ${e^{2x}} - 5{e^x} + 6 = 0$ are $x = \log 3$ and $x = \log 2$.

Note: Quadratic equations are the polynomial equations with degree of the variable or unknown as $2$. Quadratic equations can be solved by splitting the middle term, using the quadratic formula and completing the square method. Some equations don’t appear to be quadratic but can be reduced into quadratic equations using substitution.