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(4 / (3 - sqrt(5))) ^ 2 - ((6 - 5 * sqrt(6)) / (5 - sqrt(6))) ^ 2 = 2 * sqrt(61 + 24*sqrt(5))

$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$

How to prove it is right equality?

I come up with $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$, but still can't get it to the obvious equality.

Any ideas?

  • 1
    @hey, why don't you put as many close brackets as open brackets? I added them so that the question makes sense.2010-10-03
  • 6
    Please write something descriptive in the title of a question.2010-10-03
  • 0
    @muad: nice edit, thanks.2010-10-03
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    LaTeX conversion of (16 / (14 - 6*sqrt(5))) - 6 = 2 * sqrt(61 + 24*sqrt(5)) : $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$2010-10-03
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    @hey, Suggestion: change title to "How to prove an algebraic numerical equality with radicals?"2010-10-03
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    Technically it is an equality, not an identity, because there is no variables such as $(a+b)^2=a^2+2ab+b^2$2010-10-04

5 Answers 5

5

Simplify both sides to $8+6\sqrt{5}$, as they are both equal to this.

Just square the RHS to see this and rationalise the denominators on the LHS.

After the rationalisation of the denominators on the LHS (which is very quick) you obtain $(3+\sqrt{5})^2-6,$ and hence the $8+6\sqrt{5}.$

3

Since

$$\dfrac{4^{2}}{\left( 3-\sqrt{5}\right) ^{2}}-\dfrac{\left( 6-5\sqrt{6}% \right) ^{2}}{\left( 5-\sqrt{6}\right) ^{2}}=\dfrac{4^{2}\left( 5-\sqrt{6}% \right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}}{% \left( 3-\sqrt{5}\right) ^{2}\left( 5-\sqrt{6}\right) ^{2}}$$

and

$$\dfrac{1}{\left( 3-\sqrt{5}\right) ^{2}\left( 5-\sqrt{6}\right) ^{2}}=\dfrac{% \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}}{5776},$$

to prove

$$\left( \dfrac{4}{3-\sqrt{5}}\right) ^{2}-\dfrac{\left( 6-5\sqrt{6}\right) ^{2}}{\left( 5-\sqrt{6}\right) ^{2}}=2\sqrt{61+24\sqrt{5}}\quad (1)$$

it is enough to show that

$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}% \right) ^{2}$

$=11552\sqrt{61+24\sqrt{5}}$.

But (see addendum)

$$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}% \right) ^{2}=3656\sqrt{5}+46208.$$

It remains to prove

$$\left( 34656\sqrt{5}+46208\right) ^{2}=11552^{2}\left( 61+24\sqrt{5}% \right) \quad (2).$$

The left hand side is

$$\left( 3656\sqrt{5}+46208\right) ^{2}=3202768896\sqrt{5}+8140370944$$

and the right hand side is equal:

$$11552^{2}\left( 61+24\sqrt{5}\right) =3202768896\sqrt{5}+8140370944,$$

which proves your equality.


Addendum:

$$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}$$

$$=16\left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}\left( 5-\sqrt{% 6}\right) ^{2}-\left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}$$

$$=5776\left( 3+\sqrt{5}\right) ^{2}-34656=34656\sqrt{5}+46208$$


Addendum 2: The second equality stated in the question

$$\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}\quad (3)$$

is equivalent to

$$\dfrac{8}{14-6\sqrt{5}}-3=\sqrt{61+24\sqrt{5}}=\sqrt{61+\sqrt{2880}}.$$

The LHS can be transformed into

$$\dfrac{8}{14-6\sqrt{5}}-3=\dfrac{18\sqrt{5}-34}{14-6\sqrt{5}}$$

$$=\dfrac{18\sqrt{5}-34}{14-6\sqrt{5}}\times \dfrac{14+6\sqrt{5}}{14+6\sqrt{5}% }=\dfrac{64+48\sqrt{5}}{16}=4+3\sqrt{5}.$$

Then we have to prove

$$\sqrt{61+\sqrt{2880}}=4+3\sqrt{5}.\quad (4)$$

Now we apply to the LHS the following general transformation involving radicals:

$$\sqrt{A+\sqrt{B}}=\sqrt{\dfrac{A+\sqrt{A^{2}-B}}{2}}+\sqrt{\dfrac{A-\sqrt{% A^{2}-B}}{2}}$$

If $A=61,B=2880$, then $\sqrt{A^{2}-B}=\sqrt{61^{2}-2880}=29$ and

$$\sqrt{\dfrac{A+\sqrt{A^{2}-B}}{2}}=3\sqrt{5},$$

$$\sqrt{\dfrac{A-\sqrt{A^{2}-B}}{2}}=4,$$

which completes the proof.

  • 1
    You're working too hard, Américo: simplify before you take all those squares! If we write the equation as a^2 - b^2 = 2Sqrt(c), begin by finding (as Derek Jennings suggests) that a = 3 + Sqrt(5) and b = -Sqrt(6). You should also suspect that c itself is a perfect square of the form x + y Sqrt(5); you easily find x = 4 and y = 3.2010-10-03
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    whuber: I made an alternative "standard" algebraic computation (without any clever trick).2010-10-03
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    @Américo Tavares: Just to be clear, my proposal is not a "clever trick;" it's a standard way to proceed. One is guided in such problems by seeking simplification early and often. This tends to limit the work that must be done. In fact, you could halve the length of your addendum by following this principle: simplify the arguments of the squares (and of the square root) *before* performing the squaring.2010-10-04
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    @whuber: I agree with you that seeking **simplification early and often is better than rarely and later**. I also tend to do it, as opposed to what I did here. I am from a generation and a country where algebraic manipulation was very stressed, and we learned (as a problem) transformations of radicals such as the one I wrote in Addendum 2, no longer taught. I know that quite often all these manipulations are seen as boring and tedious, apart from being very time consuming. Also in trigonometry we studied many more formulae than now. In the future I will try more concise approaches.2010-10-04
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    @whuber, only a final comment: The OP did not select my answer, which is a clear evidence to me that my answer was not good.2010-10-04
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    @Américo: I wouldn't take that to be the case in general. Only one answer can be selected and you're free to disagree with the OP! The important thing is that one usually learns more by risking an answer, even if it can be improved or is wrong, than by not answering at all.2010-10-04
2

You want to show:

$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$

First multiply out the squares

$$\frac{16}{14-6\sqrt{5}} - \frac{186-60\sqrt{6}}{31-10\sqrt{6}} = 2\sqrt{61+24\sqrt{5}}$$

Then simplify it

$$\frac{16}{14-6\sqrt{5}} - 6 = 2\sqrt{61+24\sqrt{5}}$$

now we can start getting rid of the square roots on the right hand side by squaring both sides

$$\frac{256}{376-168\sqrt{5}} - \frac{192}{14-6\sqrt{5}} + 36 = 244 + 96 \sqrt{5}$$

at this point it makes sense to rationalize the denominators of the fractions

$$\left(\frac{256}{376-168\sqrt{5}}\right)\left(\frac{376+168\sqrt{5}}{376+168\sqrt{5}}\right) - \left(\frac{192}{14-6\sqrt{5}}\right)\left(\frac{14+6\sqrt{5}}{14+6\sqrt{5}}\right) + 36 = 244 + 96 \sqrt{5}$$

which simplifies to

$$376 + 168\sqrt{5} - 168 - 72\sqrt{5} + 36 = 244 + 96 \sqrt{5}$$

collecting like terms now gives

$$(376 - 168 + 36) + (168 - 72)\sqrt{5} = 244 + 96 \sqrt{5}$$

which is easily seem to be equal.


Another approach is, using the fact that algebraic numbers cannot be too close together - compute the first few digits of both sides of the original identity and compare them.

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    You probably want to make sure both sides have the same sign before you square them.2010-10-03
1

Specialize $\rm\ \ \ b=3,\ \ c = 5,\ \ d = 6 \ \Rightarrow\ a = 4\ \ $ in this simple derivation:

$\rm\quad\quad\quad\quad\displaystyle \bigg(\frac{b^2-c}{b-\sqrt{c}}\bigg)^2 - \bigg(\frac{d -5\sqrt{d}}{5-\sqrt d}\bigg)^2$

$\rm\quad\quad =\quad \:(\: b \ \: + \: \sqrt{c}\ )^{\:2} \ \ \:-\ \ \ (\:-\:\sqrt{d}\:)^{2} $

$\rm\quad\quad =\ \ 2\ (\:a + b \sqrt{c}\ )\:, \quad 2\ a\ =\ b^2+c-d $

$\rm\quad\quad =\ \ 2\:\sqrt{a^2+b^2\:c+2\:a\:b\sqrt{c} } $

NOTE $\:$ Replacing numbers by functions makes the proof both simpler and more general - similar to your recently asked question

0

Trivial. Lets simplify the left part: multiply first fraction to the sum of expressions in its denominator, and get whole part in second fraction and do with second fraction the same that we did with first:

$(\frac{4(3+\sqrt{5})}{9-5})^2 - (\frac{25-19 - \sqrt{6}}{5-\sqrt{6}}) = (3+\sqrt{5})^2 - (5-\frac{19(5+\sqrt{6})}{25-6})^2 = (3+\sqrt{5})^2 - 6 = 8 + 6\sqrt{5}$

Now lets prove this:

$8+6\sqrt{5}=2\sqrt{61+24\sqrt{5}}$

divide by two and square:

$4+3\sqrt{5} = \sqrt{61+24\sqrt{5}}$

$16+45+24\sqrt{5}=61+24\sqrt{5}$

Thats all.