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

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sauravshakyaBest ResponseYou've already chosen the best response.0
Hint: (a+b) (c+d) = ac +ad + bc +bd
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
do i keep the roots?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
yes first keep the roots.... tell me what u got
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
ok so I got \[\sqrt{x}\sqrt{x}+\sqrt{x}\sqrt{27}+\sqrt{3}\sqrt{x}+\sqrt{3}\sqrt{27}\]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
dw:1356401518093:dwNow,
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
\[x+\sqrt{27x}+\sqrt{3x}+9\]
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
or i guess\[x+\sqrt{30x}+9\]
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
I figured out the answer is \[x+4\sqrt{3x}+9\] But I do not understand how you get \[4\sqrt{3x}\] from \[\sqrt{27x}+\sqrt{3x}\]
 one year ago

calculusfunctionsBest 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!
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
So if \[\sqrt{27x}= 3\sqrt{3}\] you still have the other \[3\sqrt{3}\] don't you?
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
I'm sorry, add an x into those above.
 one year ago

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

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

kirk.freedmanBest ResponseYou've already chosen the best response.1
oh wait so rt27x= 3rt3x, and still have the rt3x though??
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
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 \[3\sqrt{3x}+\sqrt{3x}\]
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
Did you not read every single line of my explanation above, because I explained this very clearly.
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
OHHHHHHHHH I got it, thanks so much!
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
3 root 3 + 1 root 3 = 4 root 3 because 3 + 1 = 4.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
Good, Excellent!
 one year ago

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