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

#### nanase

##### Junior Member
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

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

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
Vague hint: Turn this into a perfect square.

[imath]\sqrt{48} + 13= 4\sqrt{3}+13=?[/imath]

Yes, I made it to $(2+\sqrt{3})^{2}$Then I obtain $\frac{\sqrt{3+\sqrt{3+\sqrt{3}}}}{\sqrt{2}(\sqrt{3}+1)}$I am not sure what I can cancel out here?
The repeating root 3 on top looks suspicious though..

Yes, I made it to $(2+\sqrt{3})^{2}$Then I obtain $\frac{\sqrt{3+\sqrt{3+\sqrt{3}}}}{\sqrt{2}(\sqrt{3}+1)}$I am not sure what I can cancel out here?
The repeating root 3 on top looks suspicious though..
$(2+\sqrt{3})^2 \neq 13 + 4\sqrt{3}$

Vague hint: Turn this into a perfect square.

[imath]\sqrt{48} + 13= 4\sqrt{3}+13=?[/imath]
sorry I made careless mistakes above
a new try here $13+4\sqrt{3} = 6+7+4\sqrt{3} = 6 + (2+\sqrt{3})^{2}$

sorry I made careless mistakes above
a new try here $13+4\sqrt{3} = 6+7+4\sqrt{3} = 6 + (2+\sqrt{3})^{2}$
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?
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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)

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?
Do it again for [imath]4-2\sqrt{3}[/imath].

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)
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].

Do it again for [imath]4-2\sqrt{3}[/imath].
[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?

[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?
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]

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]
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 $\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}(1+\sqrt{3})}$perfect square again BBB?

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 $\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}(1+\sqrt{3})}$
Halfway there. Leave the denominator as is and multiply by its conjugate.

$$\displaystyle \frac{\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$$

$\frac{\sqrt{12+6\sqrt{3}} - \sqrt{4+2\sqrt{3}}}{4}$
I got that, how can I simplify the numerator?

$\frac{\sqrt{12+6\sqrt{3}} - \sqrt{4+2\sqrt{3}}}{4}$
I got that, how can I simplify the numerator?
Perfect squares.

Perfect squares.
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
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all simplified to 1/2
Thank you so much for bearing & helping me thru this. Appreciate it!

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sorry I made careless mistakes above
a new try here $13+4\sqrt{3} = 6+7+4\sqrt{3} = 6 + (2+\sqrt{3})^{2}$
Try (a+bsqrt(3))2 = a2 + 3b2 + 2absqrt(3).

So a2 + 3b2 =13 and 2ab=4. Now solve for a and b

Try (a+bsqrt(3))2 = a2 + 3b2 + 2absqrt(3).

So a2 + 3b2 =13 and 2ab=4. Now solve for a and b
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 $(3x-1)(x-4)=0$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?

In case it interests anyone, here is an article that includes a simple way to handle these radicals:

"Easy Method 1" nicely simplifies every radical in this problem.