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Question

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

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Solution

## Find the number of integral values of $x$:The given expression is $\sqrt{-{x}^{2}+10x-16}.Simplify the following expression:$\sqrt{-{x}^{2}+10x-16}Take square on both sides of $\sqrt{-{x}^{2}+10x-16}:$⇒-{x}^{2}+10x-16<{x}^{2}-4x+4\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-14x+20>0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-7x+10>0\phantom{\rule{0ex}{0ex}}⇒\left(x-5\right)\left(x-2\right)>0\phantom{\rule{0ex}{0ex}}⇒\left(x-5\right)>0,\mathrm{or}\left(x-2\right)>0\phantom{\rule{0ex}{0ex}}⇒x>5,\mathrm{or}x<2\phantom{\rule{0ex}{0ex}}\therefore \mathrm{x}\in \left(-\infty ,2\right)\mathrm{U}\left(5,\infty \right)\dots \left(2\right)$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 $6,7,8$.

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