anonymous
  • anonymous
I need help with simplifying
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[6a \sqrt{80b}-\sqrt{180a ^{2}b}+5a \sqrt{b}\]
Jhannybean
  • Jhannybean
Look at the middle term and take the prime factorization of 180
Jhannybean
  • Jhannybean
Do the same for 80

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More answers

anonymous
  • anonymous
60 & 2 for 180 then 40&2 for 80 right?
Jhannybean
  • Jhannybean
\[6a\sqrt{20\cdot 2 \cdot 2 \cdot b}- \sqrt{20 \cdot 9 \cdot a\cdot a \cdot b}+5a\sqrt{b}\]
Jhannybean
  • Jhannybean
\[60 \cdot 2 \ne 180\]
anonymous
  • anonymous
lol i had meant 90
Jhannybean
  • Jhannybean
Yes that works too :)
Jhannybean
  • Jhannybean
now factor both 20's and the 9 into simpler, prime factors.
anonymous
  • anonymous
10 x 2 for 20's & 3 x 3 for 9
Jhannybean
  • Jhannybean
Good. \[6a\sqrt{10\cdot 2\cdot 2 \cdot 2 \cdot b}- \sqrt{10 \cdot 2 \cdot 3\cdot 3 \cdot a\cdot a \cdot b}+5a\sqrt{b}\]
Jhannybean
  • Jhannybean
factor both 10's one more time into prime factors.
anonymous
  • anonymous
5 & 2
Jhannybean
  • Jhannybean
Great \(\checkmark\) \[6a\sqrt{5\cdot 2\cdot 2\cdot 2 \cdot 2 \cdot b}- \sqrt{5\cdot 2 \cdot 2 \cdot 3\cdot 3 \cdot a\cdot a \cdot b}+5a\sqrt{b}\]
Jhannybean
  • Jhannybean
Now for every fair of 2 numbers, 1 of them will be extracted from the square root and multiplied to the term outside the square root.
Jhannybean
  • Jhannybean
pair*
Jhannybean
  • Jhannybean
For example, to start you off. \[6a\sqrt{5\color{blue}{\cdot 2\cdot 2}\color{red}{\cdot 2 \cdot 2} \cdot b} = 6a \cdot \color{blue}{2} \cdot \color{red}{2}\sqrt{5 \cdot b}\]
anonymous
  • anonymous
\[48a \sqrt{5\times2timesb}-6a \sqrt{5\times2\times3 \times a \times b}+5a \sqrt{b}\]
Jhannybean
  • Jhannybean
Not quite. In the first part you still have a 2 left within the square root when its been paired off and pulled out already. You should just have a 5 and a b left within the square root there since they are lone pairs and cannot be pulled out. Once things are paired off, they are not rewritten within the square root. Second part, the 6a looks right, therefore the only thing that should be left within the square root should be 5 and b.
anonymous
  • anonymous
ohh okay i think i got it now
Jhannybean
  • Jhannybean
It should look like this: \[6a\sqrt{5\color{blue}{\cdot 2\cdot 2}\color{red}{\cdot 2 \cdot 2} \cdot b} = 6a \cdot \color{blue}{2} \cdot \color{red}{2}\sqrt{5 \cdot b} = 24a\sqrt{5b}\]\[- \sqrt{5\cdot\color{red}{ 2 \cdot 2} \cdot \color{blue}{3\cdot 3} \cdot \color{green}{a\cdot a} \cdot b} = -6a\sqrt{5b}\]
anonymous
  • anonymous
yeah thats what i got
anonymous
  • anonymous
|dw:1441173348642:dw|
Jhannybean
  • Jhannybean
\[(24a-6a)\sqrt{5b} +5a\sqrt{b}= 18a\sqrt{5b}+5a\sqrt{b} \]
Jhannybean
  • Jhannybean
Coool, good job!
anonymous
  • anonymous
yup, and thank you for your help
Jhannybean
  • Jhannybean
Np.

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