simplifying another surds: (sqrt(3 + sqrt(5 - sqrt(13 + sqrt(48))))) / (sqrt(6) + sqrt(2)

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nanase

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I want to simplify this surds which will be equal to 1/2 on my answer, but kinda stuck on how to start the process
surds.jpg
I have two plans in mind :
1. by multiplying with conjugates, but I think it will mess up the numerator
2. I can still change root 48 into 4 root 3, but not sure what to do next?

can I get a hint or some opening steps on how to tackle this one?
Thank you
 
nanaseailie said:
I want to simplify this surds which will be equal to 1/2 on my answer, but kinda stuck on how to start the process
View attachment 37231
I have two plans in mind :
1. by multiplying with conjugates, but I think it will mess up the numerator
2. I can still change root 48 into 4 root 3, but not sure what to do next?

can I get a hint or some opening steps on how to tackle this one?
Thank you
Click to expand...
Vague hint: Turn this into a perfect square.

[imath]\sqrt{48} + 13= 4\sqrt{3}+13=?[/imath]
 
Yes, I made it to [math](2+\sqrt{3})^{2}[/math]Then I obtain [math]\frac{\sqrt{3+\sqrt{3+\sqrt{3}}}}{\sqrt{2}(\sqrt{3}+1)}[/math]I am not sure what I can cancel out here?
The repeating root 3 on top looks suspicious though..
 
nanaseailie said:
Yes, I made it to [math](2+\sqrt{3})^{2}[/math]Then I obtain [math]\frac{\sqrt{3+\sqrt{3+\sqrt{3}}}}{\sqrt{2}(\sqrt{3}+1)}[/math]I am not sure what I can cancel out here?
The repeating root 3 on top looks suspicious though..
Click to expand...
[math](2+\sqrt{3})^2 \neq 13 + 4\sqrt{3}[/math]
 
nanaseailie said:
sorry I made careless mistakes above
a new try here [math]13+4\sqrt{3} = 6+7+4\sqrt{3} = 6 + (2+\sqrt{3})^{2}[/math]
Click to expand...
You want the entire expression to be a perfect square. That's not a perfect square because of the +6.

Try [imath]13 + 4\sqrt{3} = 1 + 12 + 4\sqrt{3}[/imath]
 
I see, yes I got it [imath](1+2\sqrt{3})^{2}[/imath]
so now it becomes [imath]\frac{\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{2}(\sqrt{3}+1)}[/imath]
Do I still have to find a perfect square again?
.
.
side mission question, is there any tips on how to find perfect square quickly if it involves surds? (like how do I know to split the 13 above into 1 and 12, not 2 and 10 for example)
 
nanase said:
I see, yes I got it [imath](1+2\sqrt{3})^{2}[/imath]
so now it becomes [imath]\frac{\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{2}(\sqrt{3}+1)}[/imath]
Do I still have to find a perfect square again?
.
.
side mission question, is there any tips on how to find perfect square quickly if it involves surds?
Click to expand...
Do it again for [imath]4-2\sqrt{3}[/imath].
 
nanase said:
side mission question, is there any tips on how to find perfect square quickly if it involves surds? (like how do I know to split the 13 above into 1 and 12, not 2 and 10 for example)
Click to expand...
It is somewhat a guess with experience.

To turn [imath]4\sqrt{3} + 13[/imath] into a perfect square, you can add and subtract a chosen constant term.
First, notice that [imath](2\sqrt{3})^2=12[/imath]

So [imath]4\sqrt{3} + 13 + (2\sqrt{3})^2 = (2\sqrt{3}+c)^2[/imath]

Expand and compare coefficient gives [imath]c=1[/imath].
 
nanase said:
[imath]=(1-\sqrt{3})^{2}[/imath]

so now it becomes [imath]\frac{\sqrt{4-\sqrt{3}}}{\sqrt{2}(\sqrt{3}+1)}[/imath]
do I still have to find perfect square for numerator?
Click to expand...
Close but wrong due to the domain.

[imath]\sqrt{a^2} = a \text{ for all } a \ge 0[/imath]. So it must be [imath](\sqrt{3}-1)^2[/imath]
 
BigBeachBanana said:
Close but wrong due to the domain.

[imath]\sqrt{a^2} = a \text{ for all } a \ge 0[/imath]. So it must be [imath](\sqrt{3}-1)^2[/imath]
Click to expand...
hmm interesting for me that domain plays a factor here, and yes I will switch to [imath](\sqrt{3}-1)^{2}[/imath]

so I end up with final expression [math]\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}(1+\sqrt{3})}[/math]perfect square again BBB?
 
nanase said:
hmm interesting for me that domain plays a factor here, and yes I will switch to [imath](\sqrt{3}-1)^{2}[/imath]

so I end up with final expression [math]\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}(1+\sqrt{3})}[/math]
Click to expand...
Halfway there. Leave the denominator as is and multiply by its conjugate.

\(\displaystyle \frac{\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}\)
 
[math]\frac{\sqrt{12+6\sqrt{3}} - \sqrt{4+2\sqrt{3}}}{4}[/math]
I got that, how can I simplify the numerator?
 
BigBeachBanana said:
Perfect squares.
Click to expand...
Hi BBB, I can do the one on the right which is equal to [imath](1+\sqrt{3})^{2}[/imath]
and the left is equal to [imath](3+\sqrt{3})^{2}[/imath]
yippeee
.
.
all simplified to 1/2
Thank you so much for bearing & helping me thru this. Appreciate it!
 
Last edited:
nanase said:
sorry I made careless mistakes above
a new try here [math]13+4\sqrt{3} = 6+7+4\sqrt{3} = 6 + (2+\sqrt{3})^{2}[/math]
Click to expand...
Try (a+bsqrt(3))2 = a2 + 3b2 + 2absqrt(3).

So a2 + 3b2 =13 and 2ab=4. Now solve for a and b
 
Steven G said:
Try (a+bsqrt(3))2 = a2 + 3b2 + 2absqrt(3).

So a2 + 3b2 =13 and 2ab=4. Now solve for a and b
Click to expand...
ah thank you Steven G, this is the systematic way!
But in order to apply this, I should've realised that [imath](13+4\sqrt{3})=(a+b\sqrt{3})^{2}[/imath] yes?
another question I have after trying out your algebraic suggestion, I obtained [imath]3b^{4}-13b^{2}+4=0[/imath]
then I simplified into [imath]3x^{2}-13x+4=0[/imath] for solving which gives me [math](3x-1)(x-4)=0[/math]Thus x=4 and x=1/3
but x = [imath]b^{2}[/imath]
so [imath]b^{2}=4[/imath] and [imath]b^{2}=\frac{1}{3}[/imath]
Therefore b = 2 and b = [imath]\frac{1}{\sqrt{3}}[/imath]
There are 2 answers for b, how do we know for sure that b=2 is the one we use?
 
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