Help me please
screencap in comments.
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- Jravenv

Help me please
screencap in comments.
best answer rewarded

- katieb

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- Jravenv

##### 1 Attachment

- anonymous

1. simplify square roots to simplify the expression more:
\[((\sqrt{25}*\sqrt{3a})+(\sqrt{4}*\sqrt{3a})-(\sqrt{9}*\sqrt{3a}))/3a\]

- KyanTheDoodle

Probably gonna get banned for saying this, but every time I read your question, I see "best answer retarded"

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- anonymous

2. Next solve the square roots:
\[((5*\sqrt{3a})+(2*\sqrt{3a})-(3*\sqrt{3a})/\sqrt{3a}\]

- anonymous

Note: first step denominator should be square root 3a
3. the numerator has like terms so:
\[(5+2-3)(\sqrt{3a})/(\sqrt{3a})\rightarrow 5+2-3\rightarrow 4\]

- UsukiDoll

as I mentioned plenty times before...there are different methods to solving the problem in Mathematics as long as it doesn't break the Math rules.

- UsukiDoll

but.. @hughfuve your answer isn't in the choices given... so something is wrong.

- UsukiDoll

because ... (I hope I still remember this right) when you're changing to exponent form \[\sqrt{3a} \rightarrow (3a)^{\frac{1}{2}}\]

- UsukiDoll

you're distributing that 1/2 all over

- UsukiDoll

Ok. I know what's going on.. we need a \[\sqrt{3a}\] in the numerator so we can cancel out the \[\sqrt{3a}\] denominator

- UsukiDoll

so we need to split up 75, 27, and 12

- UsukiDoll

\[\frac{\sqrt{75a}+\sqrt{12a}-\sqrt{27a}}{\sqrt{3a}}\]

- UsukiDoll

ok this is going to work :)
we want that denominator to leave but before we have that we need to split up 75, 12, and 27.. we need perfect square numbers

- UsukiDoll

is @Jravenv here right now?
I want to guide her through this process
we need to know
what perfect squares can we use so we can take the square root
so what is 3 x ? = 75, 3 x ? = 12, and 3 x ? = 27?
there are 3 perfect square numbers

- UsukiDoll

another way is what's 75/3, 12/4 , and 27/3 ???

- UsukiDoll

sorry 75/3, 12/3, and 27/3

- Jravenv

114/3

- UsukiDoll

this is separate
what is 75/3
or 3 x ? = 75

- UsukiDoll

we're not adding those three fractions.. we are using division for
75/3 =?
12/3 =?
27/3 = ?

- Jravenv

25

- UsukiDoll

yes! so what about 12/3 = ? and 27/3 = ?

- Jravenv

4 and 8 and a 3rd

- UsukiDoll

what is 12 divided by 3?
what is 27 divided by 3?

- Jravenv

4 and 9

- UsukiDoll

\[\frac{\sqrt{25 \cdot 3 a}+\sqrt{4 \cdot 3a}-\sqrt{9 \cdot 3 a}}{\sqrt{3a}}\]

- UsukiDoll

so all we need is the square root of 25, 4, and 9
what is the square root of 25?
what is the square root of 4?
what is the square root of 9?

- Jravenv

5 2 and 3?

- UsukiDoll

yes.

- UsukiDoll

\[\frac{5\sqrt{ 3 a}+2\sqrt{ 3a}-3\sqrt{ 3 a}}{\sqrt{3a}}\]

- UsukiDoll

now we noticed that there is a term in common and I want to factor it out because i want to get rid of the denominator.. so what can we pull out of the numerator?

- UsukiDoll

each term in the numerator has it, so it's safe to factor out :)

- UsukiDoll

I can't factor 5 2 and 3 out, but I can factor _____________?

- Jravenv

3a

- UsukiDoll

\[\frac{\sqrt{ 3 a}(5+2-3)}{\sqrt{3a}}\]
yes notice that now we can get rid of the square root ?

- UsukiDoll

we're left with a very simple equation... using order of operations PEMDAS... well just AS...we just add and subtract.

- Jravenv

So its 4

- UsukiDoll

yes it's 4.
5+2-3
using order of operations..addition first
5+2 = 7
7-3 (subtraction)
7-3 = 4

- anonymous

geeez I completely forgot how to do that.. you are quite the doll usuki.

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