Powers and roots question

bumblebee123

Junior Member
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Jan 3, 2018
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200
Hi, I 've been attempting to figure this question out for so long but I just can't work it out!

The question is: Show that ( (1-sqrt3)^2 ) / sqrt3 = -2 + ( ( 4sqrt3 )/ 3 )

I've tried rationalizing the denominator and managed to get 2/3

I have no idea what to do. Can someone please explain? any help would be really appreciated! :)
 
There seems to be a typo:
I'm getting \(\displaystyle \dfrac{(1 - \sqrt{3} ) ^2}{ \sqrt{3} } = 2 - \dfrac{4 \sqrt{3} }{3} \)

Try starting by expanding the LHS: \(\displaystyle (1 - \sqrt{3} )^2 = 1 - 2 \sqrt{3} + 3\).

Can you finish?

-Dan
 
There seems to be a typo:
I'm getting \(\displaystyle \dfrac{(1 - \sqrt{3} ) ^2}{ \sqrt{3} } = 2 - \dfrac{4 \sqrt{3} }{3} \)

Try starting by expanding the LHS: \(\displaystyle (1 - \sqrt{3} )^2 = 1 - 2 \sqrt{3} + 3\).

Can you finish?

-Dan
No I don't think there is a typo??
 
( ? I forgot to distribute the -1 in my solution. ) I edited it out of my post.

Thanks for the catch!

-Dan
 
the question is: show that ( (1-sqrt3)^2 ) / sqrt3 = -2 + ( ( 4sqrt3 )/ 3 )

it's not: ( (1-sqrt3)^2 ) / sqrt3 = 2 - ( ( 4sqrt3 )/ 3 )

I've tried expanding the numerator and have got: ( 4 - 2 sqrt3 ) / sprt3

I can't follow your solutions, can you explain more? thank you :)
 
the question is: show that ( (1-sqrt3)^2 ) / sqrt3 = -2 + ( ( 4sqrt3 )/ 3 )
it's not: ( (1-sqrt3)^2 ) / sqrt3 = 2 - ( ( 4sqrt3 )/ 3 )
I've tried expanding the numerator and have got: ( 4 - 2 sqrt3 ) / sprt3
I can't follow your solutions, can you explain more?
\(\displaystyle \begin{align*}\dfrac{(1-\sqrt3)^2}{\sqrt3}&=\dfrac{1-2\sqrt3+3}{\sqrt3} \\&=\dfrac{4-2\sqrt3}{\sqrt3}\\&=\dfrac{4}{\sqrt3}-2\\&=\dfrac{4\sqrt3}{3}-2 \end{align*}\)
 
Thank you for your help. I follow the expansion but how did you get rid of the sqrt3 on the 3rd line to just be left with a -2?
 
Thank you for your help. I follow the expansion but how did you get rid of the sqrt3 on the 3rd line to just be left with a -2?
Do you have the necessary background/preparation to do this material?
\(\displaystyle -\frac{2\sqrt3}{\sqrt3}=-2\), that is simple middle school material \(\displaystyle \frac{A\cdot B}{B}=\frac{A\cdot \cancel{B}}{\cancel{B}}=A\)
 
yes, but I still don't understand how (4 - 2 sqrt3 ) / sqrt3 can be turned into (4/ sqrt3 ) - 2 ?
 
yes, but I still don't understand how (4 - 2 sqrt3 ) / sqrt3 can be turned into (4/ sqrt3 ) - 2 ?
\(\displaystyle \frac{A-2B}{B}=\frac{A}{B}-\frac{2B}{B}=\frac{A}{B}-\frac{2\cancel{B}}{\cancel{B}}=\frac{A}{B}-2\)
 
Thank you for your help. I follow the expansion but how did you get rid of the sqrt3 on the 3rd line to just be left with a -2?

Just to check that I'm understanding you correctly: The step you're confused about is this one, yes?

\(\displaystyle \frac{4-2\sqrt{3}}{\sqrt{3}} = \frac{4}{\sqrt{3}} - 2\)

If yes, then let's look at some simpler examples to see if we can gain some insight. Suppose we had:

\(\displaystyle \frac{3 + 14}{7}\)

Now, obviously, we can simply add everything up and see that's the same as \(\frac{17}{7}\). But we can also see that it must be equal to \(\frac{3}{7} + \frac{14}{7} = \frac{3}{7} + 2\). Now suppose we had:

\(\displaystyle \frac{5 - 18}{6}\)

In the same manner as before, we can see this is the same as \(\frac{5}{6} - \frac{18}{6} = \frac{5}{6} - 3\). If you followed those examples, then we can make the transition to your actual problem. We may not know the exact value of \(2\sqrt{3}\), but we definitely know it's a number. So we can treat it exactly the same as we would any other number. That gives us an implicit step not shown in pka's work (presumably because they assumed you would pick up on it):

\(\displaystyle \frac{4-2\sqrt{3}}{\sqrt{3}} = \frac{4}{\sqrt{3}} - \frac{2\sqrt{3}}{\sqrt{3}} = \frac{4}{\sqrt{3}} - 2\)
 
Just to check that I'm understanding you correctly: The step you're confused about is this one, yes?

\(\displaystyle \frac{4-2\sqrt{3}}{\sqrt{3}} = \frac{4}{\sqrt{3}} - 2\)

If yes, then let's look at some simpler examples to see if we can gain some insight. Suppose we had:

\(\displaystyle \frac{3 + 14}{7}\)

Now, obviously, we can simply add everything up and see that's the same as \(\frac{17}{7}\). But we can also see that it must be equal to \(\frac{3}{7} + \frac{14}{7} = \frac{3}{7} + 2\). Now suppose we had:

\(\displaystyle \frac{5 - 18}{6}\)

In the same manner as before, we can see this is the same as \(\frac{5}{6} - \frac{18}{6} = \frac{5}{6} - 3\). If you followed those examples, then we can make the transition to your actual problem. We may not know the exact value of \(2\sqrt{3}\), but we definitely know it's a number. So we can treat it exactly the same as we would any other number. That gives us an implicit step not shown in pka's work (presumably because they assumed you would pick up on it):

\(\displaystyle \frac{4-2\sqrt{3}}{\sqrt{3}} = \frac{4}{\sqrt{3}} - \frac{2\sqrt{3}}{\sqrt{3}} = \frac{4}{\sqrt{3}} - 2\)

thank you so much! yes, this was the part I was so confused about, I didn't think about separating it into two separate fractions, that's why I couldn't link the two steps together. This makes a lot more sense :)
 
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