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

[nanase]

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

 

[BigBeachBanana]

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.

$\sqrt{48} + 13= 4\sqrt{3}+13=?$

 

[nanase]

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..

 

[BigBeachBanana]

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..

$$(2+\sqrt{3})^2 \neq 13 + 4\sqrt{3}$$

 

[nanase]

Vague hint: Turn this into a perfect square.

$\sqrt{48} + 13= 4\sqrt{3}+13=?$

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

 

[BigBeachBanana]

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 $13 + 4\sqrt{3} = 1 + 12 + 4\sqrt{3}$

 

[nanase]

I see, yes I got it $(1+2\sqrt{3})^{2}$
so now it becomes $\frac{\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{2}(\sqrt{3}+1)}$
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)

 

[BigBeachBanana]

I see, yes I got it $(1+2\sqrt{3})^{2}$
so now it becomes $\frac{\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{2}(\sqrt{3}+1)}$
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 $4-2\sqrt{3}$.

 

[BigBeachBanana]

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

So $4\sqrt{3} + 13 + (2\sqrt{3})^2 = (2\sqrt{3}+c)^2$

Expand and compare coefficient gives $c=1$.

 

[nanase]

Do it again for $4-2\sqrt{3}$.

$=(1-\sqrt{3})^{2}$

so now it becomes $\frac{\sqrt{4-\sqrt{3}}}{\sqrt{2}(\sqrt{3}+1)}$
do I still have to find perfect square for numerator?

 

[BigBeachBanana]

$=(1-\sqrt{3})^{2}$

so now it becomes $\frac{\sqrt{4-\sqrt{3}}}{\sqrt{2}(\sqrt{3}+1)}$
do I still have to find perfect square for numerator?

Close but wrong due to the domain.

$\sqrt{a^2} = a \text{ for all } a \ge 0$. So it must be $(\sqrt{3}-1)^2$

 

[nanase]

Close but wrong due to the domain.

$\sqrt{a^2} = a \text{ for all } a \ge 0$. So it must be $(\sqrt{3}-1)^2$

hmm interesting for me that domain plays a factor here, and yes I will switch to $(\sqrt{3}-1)^{2}$

so I end up with final expression $$\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}(1+\sqrt{3})}$$perfect square again BBB?

 

[BigBeachBanana]

hmm interesting for me that domain plays a factor here, and yes I will switch to $(\sqrt{3}-1)^{2}$

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.

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

 

[nanase]

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

 

[BigBeachBanana]

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

Perfect squares.

 

[nanase]

Perfect squares.

Hi BBB, I can do the one on the right which is equal to $(1+\sqrt{3})^{2}$
and the left is equal to $(3+\sqrt{3})^{2}$
yippeee
.
.
all simplified to 1/2
Thank you so much for bearing & helping me thru this. Appreciate it!

 

[Steven G]

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

 

[nanase]

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 $(13+4\sqrt{3})=(a+b\sqrt{3})^{2}$ yes?
another question I have after trying out your algebraic suggestion, I obtained $3b^{4}-13b^{2}+4=0$
then I simplified into $3x^{2}-13x+4=0$ for solving which gives me $$(3x-1)(x-4)=0$$Thus x=4 and x=1/3
but x = $b^{2}$
so $b^{2}=4$ and $b^{2}=\frac{1}{3}$
Therefore b = 2 and b = $\frac{1}{\sqrt{3}}$
There are 2 answers for b, how do we know for sure that b=2 is the one we use?

 

[Dr.Peterson]

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

Denesting Radicals (or Unnesting Radicals)

Simplifying a square root that contains a rational number plus or minus a square root

brownmath.com

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