Simplify the following expression

[tahayassen]

Homework Statement

http://img513.imageshack.us/img513/88/71893688.png

Homework Equations

Not applicable.

The Attempt at a Solution

http://img441.imageshack.us/img441/846/14479081.png

I'm already stuck! I calculated both expressions and they don't equal each other. :(

 

[tahayassen]

Actually, don't bother helping me with this one. I think I've figured out. I needed to find the LCD between the two fractions.

Instead, answer this: why can't I take the square root of a binomial?

edit: I'm stuck again with this question. I can't seem to get rid of the large radical even after I combine the fractions. :(

 

[Mentallic]

You can't do that because generally,

$$\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$$

And if you want to see why, if these are equal, then their squares must be equal. Try squaring both sides and see what happens.

This is one of those problems that unless you know where you're going, it's going to be difficult finding your way there. I tried a few things out but I don't see the radicals being simplified. I believe the only way to simplify this expression is to take a common denominator and then take a factor of 4 in the denominator out of the larger surd.

 

[I like Serena]

Wolfram comes up with a nice one:
$$
\frac 1 2 \sqrt{2 + \sqrt{2 - \sqrt 3}}
$$

Beyond that I also believe there's not much that is useful to simplify.

 

[Mentallic]

Wolfram comes up with a nice one:
$$
\frac 1 2 \sqrt{2 + \sqrt{2 - \sqrt 3}}
$$

If you consider further nested roots as being simpler, then sure

I made a futile attempt to see if the larger surd can be removed by trying to find if there exist integers a,b,c,d to the following equality:

$$\left(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\right)^2= \sqrt{6} -\sqrt{2}+4$$

but to my amazement (not really, I was expecting it) to figure this out, one would need to solve these non-linear simultaneous equations:

$$a^2+2b^2+3c^2+6d^2=4$$
$$ab+3cd=\frac{-1}{2}$$
$$ac+2bd=0$$
$$ad+bc=\frac{1}{2}$$

And I haven't a clue how to proceed - plus I'm not really feeling up for the challenge considering its pretty unlikely that the variables are all in fact integers.

 

[I like Serena]

If you consider further nested roots as being simpler, then sure

Less work to type into a calculator.
And it looks nice!

And I haven't a clue how to proceed - plus I'm not really feeling up for the challenge considering its pretty unlikely that the variables are all in fact integers.

Neither have I, but if you do proceed, please swap the signs of the 2nd and 4th line. ;)

 

[Curious3141]

I'm pretty sure there's no way to make that surd much simpler in any meaningful sense.

 

[Mentallic]

Less work to type into a calculator.
And it looks nice!

Agreed, it looks pretty sweet!

Neither have I, but if you do proceed, please swap the signs of the 2nd and 4th line. ;)

Uhh... How I managed to transcribe that incorrectly I'll never know

 

[Ray Vickson]

If you consider further nested roots as being simpler, then sure

I made a futile attempt to see if the larger surd can be removed by trying to find if there exist integers a,b,c,d to the following equality:

$$\left(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\right)^2= \sqrt{6} -\sqrt{2}+4$$

but to my amazement (not really, I was expecting it) to figure this out, one would need to solve these non-linear simultaneous equations:

$$a^2+2b^2+3c^2+4d^2=4$$
$$ab+3cd=\frac{-1}{2}$$
$$ac+2bd=0$$
$$ad+bc=\frac{1}{2}$$

And I haven't a clue how to proceed - plus I'm not really feeling up for the challenge considering its pretty unlikely that the variables are all in fact integers.

Well, this is probably not what is wanted in high-school or first-year college, but one _can_ get a solution using Groebner bases. Here is what happens in Maple 14:
sys:={a^2+2*b^2+3*c^2+4*d^2-4,2*a*b+6*c*d+1,a*c+2*b*d,2*a*d+2*b*c-1};
with(Groebner);

B:=Basis(sys,plex(a,b,c,d));
B := [1-256*d^2+19072*d^4-626560*d^6+10231872*d^8-85606400*d^10+351143936*d^12-621502464*d^14+359817216*d^16,

58700876744*d-6772491730030*d^3+516306245937248*d^5-16676091040182256*d^7+214982693609933568*d^9-1128998897071871232*d^11+2264133466874087424*d^13-1391829359823378432*d^15+5466782401*a,

20794688424*d-25887729161344*d^3+1613081383661048*d^5-35738765248743936*d^7+355188616828494592*d^9-1606762324171226112*d^11+2991485535250916352*d^13-1775184867367059456*d^15+5466782401*b,

127738358584*d-36353540383530*d^3+2108143299164768*d^5-47032277980064368*d^7+476144844758761728*d^9-2186167800738890496*d^11+4105190085058529280*d^13-2446662083621640192*d^15+5466782401*c]

So, d is one of the roots of the 16th degree polynomial in B[1]; then we can get a, b and c by setting B[2], B[3] and B[4] to zero. See? Nothing to it!

RGV

 

[I like Serena]

Well, all roots of B[1] appear to be irrational (according to Wolfram), so I don't think we get a nice simplification. ;)

 

[SammyS]

Look at various ways to express $\displaystyle\cos\left(\frac{5\pi}{24}\right)\,,$ and/or $\displaystyle\sin\left(\frac{7\pi}{24}\right)\,.$

 

[I like Serena]

Nice! However did you find that one?

 

[SammyS]

Nice! However did you find that one?

It reminded me of a problem I used to give students in which they were to find $\displaystyle\sin\left(\frac{\pi}{12}\right)\,,$ using both angle subtraction and half angle identities.

(I then had them give the approximate numerical answer to each -- mostly so they could see that the two different looking answers were equivalent.)

So I played around with that a bit for the problem in this thread.

 

[e^(i Pi)+1=0]

I went a little nuts with this problem and got...

$\frac{(6(√6-√2)+24)\sqrt{6(√6-√2)+24}}{√48(6(√6-√2)+24)}$

Are radicals still considered nested if they're within parentheses?

edit: they would be since you could just distribute in the 6, oh well.

 

[Mentallic]

Well, this is probably not what is wanted in high-school or first-year college, but one _can_ get a solution using Groebner bases. Here is what happens in Maple 14:
sys:={a^2+2*b^2+3*c^2+4*d^2-4,2*a*b+6*c*d+1,a*c+2*b*d,2*a*d+2*b*c-1};
with(Groebner);

Could you try this again but with
sys:={a^2+2*b^2+3*c^2+6*d^2-4,2*a*b+6*c*d+1,a*c+2*b*d,2*a*d+2*b*c-1};

So, d is one of the roots of the 16th degree polynomial in B[1]; then we can get a, b and c by setting B[2], B[3] and B[4] to zero. See? Nothing to it!

What a relief!

Look at various ways to express $\displaystyle\cos\left(\frac{5\pi}{24}\right)\,,$ and/or $\displaystyle\sin\left(\frac{7\pi}{24}\right)\,.$

This is one of those problems that unless you know where you're going, it's going to be difficult finding your way there.

Ahh yes that's where I remember seeing something similar to the expression $$\frac{\sqrt{6}-\sqrt{2}}{8}$$
Thanks for reminding me.