Simplify (sqrt x + sqrt 3) (sqrt x + sqrt 27)
show work or explain please

- anonymous

Simplify (sqrt x + sqrt 3) (sqrt x + sqrt 27)
show work or explain please

- chestercat

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- anonymous

##### 1 Attachment

- anonymous

Hint: (a+b) (c+d) = ac +ad + bc +bd

- anonymous

do i keep the roots?

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## More answers

- anonymous

yes first keep the roots....
tell me what u got

- anonymous

ok so I got \[\sqrt{x}\sqrt{x}+\sqrt{x}\sqrt{27}+\sqrt{3}\sqrt{x}+\sqrt{3}\sqrt{27}\]

- anonymous

good

- anonymous

|dw:1356401518093:dw|Now,

- anonymous

Can u use it too

- anonymous

\[x+\sqrt{27x}+\sqrt{3x}+9\]

- anonymous

or i guess\[x+\sqrt{30x}+9\]

- anonymous

good

- anonymous

Now what??

- anonymous

I figured out the answer is \[x+4\sqrt{3x}+9\] But I do not understand how you get \[4\sqrt{3x}\] from \[\sqrt{27x}+\sqrt{3x}\]

- calculusfunctions

@kirk.freedman \[\sqrt{27x}+\sqrt{3x}\neq \sqrt{30x}\]Therefore your above answer is incorrect. I don't know why @sauravshakya said "good"? I know he knows better than that. Maybe he was tired or rushing.
You can only add or subtract two radicals by adding or subtracting the numbers in front iff they have the same index and the same radicand. You can multiply or divide two radicals by multiplying or dividing their radicands iff they have the same index.
The only thing wrong about your answer is the root 30x. you should simplify the sq.root of 27x first. Like this:\[\sqrt{27x}=\sqrt{9}\sqrt{3x}=3\sqrt{3x}\] Thus\[(\sqrt{x}+\sqrt{3})(\sqrt{x}+\sqrt{27})\]\[=x +\sqrt{27x}+\sqrt{3x}+\sqrt{81}\]\[=x +3\sqrt{3x}+\sqrt{3x}+9\]\[=x +4\sqrt{3x}+9\]And this @kirk.freedman is your final answer!

- anonymous

So if \[\sqrt{27x}= 3\sqrt{3}\] you still have the other \[3\sqrt{3}\] don't you?

- anonymous

I'm sorry, add an x into those above.

- calculusfunctions

No the other one is just root 3. 3 root 3 is only derived from simplifying root 27. Understand?

- calculusfunctions

Well x or no x, 3 root 3 is only derived from simplifying root 27.

- anonymous

oh wait so rt27x= 3rt3x, and still have the rt3x though??

- calculusfunctions

Yes!

- anonymous

from the equation above \[x+\sqrt{27x}+\sqrt{3x}+9\] See we still have the \[\sqrt{3}\]?? do we just get rid of it or what? Because we end up having \[3\sqrt{3x}+\sqrt{3x}\]

- calculusfunctions

Did you not read every single line of my explanation above, because I explained this very clearly.

- anonymous

OHHHHHHHHH I got it, thanks so much!

- calculusfunctions

3 root 3 + 1 root 3 = 4 root 3 because 3 + 1 = 4.

- calculusfunctions

Good, Excellent!

- anonymous

Sorry I was saying good for the first one...
And I was unable to continue as my internet stop working

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