anonymous
  • anonymous
Simplify (sqrt x + sqrt 3) (sqrt x + sqrt 27) show work or explain please
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
Hint: (a+b) (c+d) = ac +ad + bc +bd
anonymous
  • anonymous
do i keep the roots?

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anonymous
  • anonymous
yes first keep the roots.... tell me what u got
anonymous
  • anonymous
ok so I got \[\sqrt{x}\sqrt{x}+\sqrt{x}\sqrt{27}+\sqrt{3}\sqrt{x}+\sqrt{3}\sqrt{27}\]
anonymous
  • anonymous
good
anonymous
  • anonymous
|dw:1356401518093:dw|Now,
anonymous
  • anonymous
Can u use it too
anonymous
  • anonymous
\[x+\sqrt{27x}+\sqrt{3x}+9\]
anonymous
  • anonymous
or i guess\[x+\sqrt{30x}+9\]
anonymous
  • anonymous
good
anonymous
  • anonymous
Now what??
anonymous
  • 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
  • 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
  • anonymous
So if \[\sqrt{27x}= 3\sqrt{3}\] you still have the other \[3\sqrt{3}\] don't you?
anonymous
  • anonymous
I'm sorry, add an x into those above.
calculusfunctions
  • calculusfunctions
No the other one is just root 3. 3 root 3 is only derived from simplifying root 27. Understand?
calculusfunctions
  • calculusfunctions
Well x or no x, 3 root 3 is only derived from simplifying root 27.
anonymous
  • anonymous
oh wait so rt27x= 3rt3x, and still have the rt3x though??
calculusfunctions
  • calculusfunctions
Yes!
anonymous
  • 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
  • calculusfunctions
Did you not read every single line of my explanation above, because I explained this very clearly.
anonymous
  • anonymous
OHHHHHHHHH I got it, thanks so much!
calculusfunctions
  • calculusfunctions
3 root 3 + 1 root 3 = 4 root 3 because 3 + 1 = 4.
calculusfunctions
  • calculusfunctions
Good, Excellent!
anonymous
  • 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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