radicals and rational exponents
- Thread starter ree
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mmm4444bot
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Easier to text.
Hello, ree!
When multiplying radicals, rational exponents make it far easier.
Example: .\(\displaystyle \sqrt[3]{x}\cdot\sqrt[4]{x}\)
Since we cannot combine "unlike" roots (only "like" roots),
. . we must perform some severe manipulations.
. . \(\displaystyle \sqrt[3]{x}\cdot\sqrt[4]{x} \;=\;\sqrt[12]{x^4}\cdot\sqrt[12]{x^3} \;=\;\sqrt[12]{x^4\cdot x^3} \;=\;\sqrt[12]{x^7}\)
Wiith rational exponents, we have:
. . \(\displaystyle x^{\frac{1}{3}}\cdot x^{\frac{1}{4}} \:=\:x^{\frac{1}{3}+\frac{1}{4}} \:=\: x^{\frac{7}{12}\)
When multiplying radicals, rational exponents make it far easier.
Example: .\(\displaystyle \sqrt[3]{x}\cdot\sqrt[4]{x}\)
Since we cannot combine "unlike" roots (only "like" roots),
. . we must perform some severe manipulations.
. . \(\displaystyle \sqrt[3]{x}\cdot\sqrt[4]{x} \;=\;\sqrt[12]{x^4}\cdot\sqrt[12]{x^3} \;=\;\sqrt[12]{x^4\cdot x^3} \;=\;\sqrt[12]{x^7}\)
Wiith rational exponents, we have:
. . \(\displaystyle x^{\frac{1}{3}}\cdot x^{\frac{1}{4}} \:=\:x^{\frac{1}{3}+\frac{1}{4}} \:=\: x^{\frac{7}{12}\)
J
JeffM
Guest
rachael said:
Hi Rachael
In general it is a very good idea to use the newtopic button to ask a new question. If you just tack your question at the end of an old topic, people will assume that you have replied to the original question, not asked a new one.
OK It is impossible to understand your question because you did not use grouping symbols. For example, is it (2 / x) + 3 or [2 / (x + 3)]. So your question is not clear. No one here is looking at your book the way you are. And is x2 supposed to mean x[sup:2ykn7xsb]2[/sup:2ykn7xsb]? The simple and clear way to show exponentiation on this site is with a ^. So, x[sup:2ykn7xsb]2[/sup:2ykn7xsb] can be rendered x^2.
So, here is what I suggest. You re-ask your question using the newtopic button (which will make some people mad so blame me) and make sure you use the symbols required to make your question clear. OK?