Can $\displaystyle \sqrt{9-x^2}$ be simplified to $\displaystyle 3-x$ ?
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No, because $\displaystyle 9-x^2=(3-x)(3+x)$ So$\displaystyle \sqrt{9-x^2}=\sqrt{(3-x)(3+x)}$ You can even put in a number to test. Let $\displaystyle x=1$ For $\displaystyle x=1$ $\displaystyle \sqrt{9-x^2}=2\sqrt{2}$ While $\displaystyle 3-x=2$
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