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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hint: (a+b) (c+d) = ac +ad + bc +bd

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes first keep the roots.... tell me what u got

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so I got \[\sqrt{x}\sqrt{x}+\sqrt{x}\sqrt{27}+\sqrt{3}\sqrt{x}+\sqrt{3}\sqrt{27}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356401518093:dwNow,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x+\sqrt{27x}+\sqrt{3x}+9\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or i guess\[x+\sqrt{30x}+9\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.2@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
 3 years ago
Best ResponseYou've already chosen the best response.0So if \[\sqrt{27x}= 3\sqrt{3}\] you still have the other \[3\sqrt{3}\] don't you?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, add an x into those above.

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.2No the other one is just root 3. 3 root 3 is only derived from simplifying root 27. Understand?

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.2Well x or no x, 3 root 3 is only derived from simplifying root 27.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh wait so rt27x= 3rt3x, and still have the rt3x though??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0from 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
 3 years ago
Best ResponseYou've already chosen the best response.2Did you not read every single line of my explanation above, because I explained this very clearly.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OHHHHHHHHH I got it, thanks so much!

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.23 root 3 + 1 root 3 = 4 root 3 because 3 + 1 = 4.

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.2Good, Excellent!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry I was saying good for the first one... And I was unable to continue as my internet stop working
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