Question

The number of integral values of $x$ satisfying $\sqrt{-x^2+10x-16}<x-2$ is

Answer 1

Find the number of integral values of $x$:

The given expression is $\sqrt{-x^2+10x-16}<x-2$.

Simplify the following expression:
$$\begin{gathered} \sqrt{-x^2+10x-16}<x-2 \\ \Rightarrow -{\mathrm{x}}^2+10\mathrm{x}-16\ge 0 \\ \Rightarrow x^2-10x+16\le 0 \\ \Rightarrow x^2-2x-8x+16\le 0 \\ \Rightarrow (x-2)(x-8)\le 0 \\ \Rightarrow (x-2)\ge 0,\mathrm{or}(x-8)\le 0 \\ \therefore \mathrm{x}\in \left[2,8\right]\dots \left(1\right) \end{gathered}$$

Take square on both sides of $\sqrt{-x^2+10x-16}<x-2$:
$$\begin{gathered} \Rightarrow -x^2+10x-16<x^2-4x+4 \\ \Rightarrow 2x^2-14x+20>0 \\ \Rightarrow x^2-7x+10>0 \\ \Rightarrow (x-5)(x-2)>0 \\ \Rightarrow (x-5)>0,\mathrm{or}(x-2)>0 \\ \Rightarrow x>5,\mathrm{or}x<2 \\ \therefore \mathrm{x}\in \left(-\infty ,2\right)\mathrm{U}\left(5,\infty \right)\dots \left(2\right) \end{gathered}$$
The points $6,7$ and $8$ are those that lie within the interval of both $\left(1\right)$ and $\left(2\right)$.

Thus, $\mathrm{x}=6,7,8$

Therefore, the number of integral values of $x$ is $6,7,8$.