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# Help...

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Simplify the expression . Show your work.

Oct 1, 2018

#1
+1336
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Oct 1, 2018
#2
+97586
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Thanks gamesmaster

$$\sqrt{x^2}=|x| \qquad not\;\;x$$

If this expression is real then y must be positive.

Also if an expression is simple then the denominator should be rational. (no Square root on the bottom)

Although your teacher may not care about this.

$$\sqrt\frac{4x^2}{3y}\\=\frac{|2x|}{\sqrt{3y}}\\ =\frac{|2x|}{\sqrt{3y}}\times \frac{\sqrt{3y}}{\sqrt{3y}}\\ =\frac{|2x|\sqrt{3y}}{3y} \\ =\left| \frac{2x\sqrt{3y}}{3y} \right|\\~\\ = \frac{2x\sqrt{3y}}{3y} \text{ when }x\ge0\\ = \frac{-2x\sqrt{3y}}{3y} \text{ when }x<0$$

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Oct 2, 2018