Question

For what values of$x$is the expression$\sqrt{6x-x^2}$ defined as a real number?

Short answer

$\sqrt{6x-x^2}$defines a real number in the interval$(0,6)$

Step-by-step solution

Step 1. Concept of real solution in square root.

In square root the solution is real when the equation under the square root is greater than$0$

Step 2. Determining the real solution.

$\begin{array}{l}6x-x^2>0\\ x(6-x)>0\left(\mathrm{Factor}\right)\\ x>0\mathrm{and}6-x>0\\ \mathrm{or},x<0\mathrm{and}6-x<0\\ \mathrm{So},x>0\mathrm{and}x<6\\ \mathrm{or},x<0\mathrm{and}x>6\end{array}$

Now$0<x<6$

Step 3. Expressing the solution set in Interval Notation.

The solution set consists of all real number less than $6$and greater than $0$. In other words, the solution of the inequality is in the interval$(0,6)$