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Mathematics 10 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

do i keep the roots?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

good

OpenStudy (anonymous):

|dw:1356401518093:dw|Now,

OpenStudy (anonymous):

Can u use it too

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

good

OpenStudy (anonymous):

Now what??

OpenStudy (anonymous):

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

OpenStudy (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!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (calculusfunctions):

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

OpenStudy (calculusfunctions):

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

OpenStudy (anonymous):

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

OpenStudy (calculusfunctions):

Yes!

OpenStudy (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 \[3\sqrt{3x}+\sqrt{3x}\]

OpenStudy (calculusfunctions):

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

OpenStudy (anonymous):

OHHHHHHHHH I got it, thanks so much!

OpenStudy (calculusfunctions):

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

OpenStudy (calculusfunctions):

Good, Excellent!

OpenStudy (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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