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Original question was to write the fraction in its simplest form:

Question: 4r/[(√6r) + 9]

I attempted to solve it by multiplying the denominator and numerator by the conjugates of the denominator, ((√6r) - 9).

My answer:

numerator: [4r*(√6r)] - 36r denominator: 6r2 - 81

Actual answer:

numerator: [4r2*(√6)] - 36r denominator: 6r2 -81

Why was the r under the √6r removed in the numerator?

Why and how did 4r become 4r2?

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closed as unclear what you're asking by Andrei, lulu, Lord Shark the Unknown, KReiser, Shailesh Jan 3 at 8:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Well, it's not clear (to me at least) whether (√6r) means $\sqrt {6r}$ or $\sqrt 6\times r$. This should be clarified. to the question in the header: of course it doesn't. $4r\times \sqrt {6r}=4r^{3/2}\times \sqrt 6$. $\endgroup$ – lulu Jan 2 at 14:48
  • $\begingroup$ Do you mean $$\frac{4r}{\sqrt{6r}+9}$$? $\endgroup$ – Dr. Sonnhard Graubner Jan 2 at 14:50
  • $\begingroup$ Why do you get $6r^2$ in the denominator? If you consider it as $\sqrt{6r}$ surely the square is $6r$ (at least assuming $r$ positive). If it's $\sqrt{6} r$ though then it's not clear why $4 r $ times $\sqrt{6} r$ would not be $4 \sqrt{6} r^2$ when $\sqrt{6} r$ times $\sqrt{6} r$ is $6 r^2$ $\endgroup$ – quid Jan 2 at 14:54
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    $\begingroup$ Please format your question, so we can read it. Look at math.stackexchange.com/help/notation, to get an idea how to do that $\endgroup$ – Andrei Jan 2 at 15:10
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    $\begingroup$ Voting to close the question as it not clear what you are asking. As you can see from the comments, nobody is clear on what it is you are trying to write. Please edit for clarity. $\endgroup$ – lulu Jan 2 at 15:12
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If r is not inside the under-root [this is suggested by your answer in denominator] then

$\frac{4r}{(\sqrt{6}r+9)}= \frac{(4r)(\sqrt{6}r-9)}{(6r^2 -81)}=\frac{4\sqrt{6}r^2-36r}{6r^2-81}$

Which is the actual answer. You are doing mistake while calculating.

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    $\begingroup$ Please format your answer. See math.stackexchange.com/help/notation or look up MathJax $\endgroup$ – Andrei Jan 2 at 15:57
  • $\begingroup$ Thanks for the help as this is my first day of using this site $\endgroup$ – BJKShah Jan 2 at 16:25
  • $\begingroup$ To add the square root, use $\sqrt{expression}$ $\endgroup$ – Andrei Jan 2 at 16:39

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