Algebra Problem of The Day-8

[BigBeachBanana]

Simplify the following expression without a calculator:
$$\sqrt{6 + 2\sqrt{3} +2\sqrt{2} + 2\sqrt{6}} - \frac{1}{\sqrt{5-2\sqrt{6}}}$$
The answer is an integer.

 

[red12dog34]

Spoiler

It is easy to see that $$6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}=1+2+3+2\cdot 1\cdot\sqrt{2}+2\cdot 1\cdot\sqrt{3}+2\cdot\sqrt{2}\cdot\sqrt{3}=\left(1+\sqrt{2}+\sqrt{3}\right)^2$$and $$\frac{1}{\sqrt{5-2\sqrt{6}}}=\sqrt{5+2\sqrt{6}}=\sqrt{5+\sqrt{24}}$$We use the formula $$\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}, \ \text{where} \ C=\sqrt{A^2-B}$$Then $$\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+1}{2}}+\sqrt{\frac{5-1}{2}}=\sqrt{3}+\sqrt{2}$$Then $$\sqrt{(1+\sqrt{2}+\sqrt{3})^2}-(\sqrt{2}+\sqrt{3})=1+\sqrt{2}+\sqrt{3}-\sqrt{2}-\sqrt{3}=1$$

 

[Harry_the_cat]

Spoiler

$\displaystyle \frac{1}{\sqrt{5}-2\sqrt{6}}$

$\displaystyle =\frac{1}{\sqrt {{\sqrt2}^2 +{\sqrt3}^2 -2\sqrt6}}$

$\displaystyle =\frac{1}{\sqrt{(\sqrt3 - \sqrt2)^2}}$

$\displaystyle =\frac{1}{\sqrt3 - \sqrt 2}$

$\displaystyle =\sqrt3 + \sqrt 2$

Because we are told that the answer is an integer, the expression under the square root sign in the first term of the original difference should be of the form $\displaystyle (a + \sqrt3 +\sqrt2)^2$ where $\displaystyle a$ is an integer.

The value of the coefficients indicate that a will be positive and equal to 1.

Checking this:

$\displaystyle (1+\sqrt3+\sqrt2)^2$

$\displaystyle =(1+\sqrt3+\sqrt2)(1+\sqrt3+\sqrt2)$

$\displaystyle = 1 +\sqrt3 +\sqrt2 +\sqrt3 +3+\sqrt6 +\sqrt2+\sqrt6 + 2$

$\displaystyle =6+2\sqrt3+2\sqrt2+2\sqrt6$

Isn't that nice!!

So LHS $\displaystyle = (1+\sqrt3 + \sqrt2) - (\sqrt3 +\sqrt2) =1$

You're welcome!

 

[Steven G]

Spoiler

$\displaystyle \frac{1}{\sqrt{5}-2\sqrt{6}}$

$\displaystyle =\frac{1}{\sqrt {{\sqrt2}^2 +{\sqrt3}^2 -2\sqrt6}}$

$\displaystyle =\frac{1}{\sqrt{(\sqrt3 - \sqrt2)^2}}$

$\displaystyle =\frac{1}{\sqrt3 - \sqrt 2}$

$\displaystyle =\sqrt3 + \sqrt 2$

Because we are told that the answer is an integer, the expression under the square root sign in the first term of the original difference should be of the form $\displaystyle (a + \sqrt3 +\sqrt2)^2$ where $\displaystyle a$ is an integer.

The value of the coefficients indicate that a will be positive and equal to 1.

Checking this:

$\displaystyle (1+\sqrt3+\sqrt2)^2$

$\displaystyle =(1+\sqrt3+\sqrt2)(1+\sqrt3+\sqrt2)$

$\displaystyle = 1 +\sqrt3 +\sqrt2 +\sqrt3 +3+\sqrt6 +\sqrt2+\sqrt6 + 2$

$\displaystyle =6+2\sqrt3+2\sqrt2+2\sqrt6$

Isn't that nice!!

So LHS $\displaystyle = (1+\sqrt3 + \sqrt2) - (\sqrt3 +\sqrt2) =1$

You're welcome!

That is so nice how you broke up the 5. Everything flowed from there. Excellent work! Even Gauss would be impressed.

 

[lookagain]

Spoiler

It is easy to see that $$6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}=1+2+3+2\cdot 1\cdot\sqrt{2}+2\cdot 1\cdot\sqrt{3}+2\cdot\sqrt{2}\cdot\sqrt{3}=\left(1+\sqrt{2}+\sqrt{3}\right)^2$$and $$\frac{1}{\sqrt{5-2\sqrt{6}}}=\sqrt{5+2\sqrt{6}}=\sqrt{5+\sqrt{24}}$$We use the formula $$\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}, \ \text{where} \ C=\sqrt{A^2-B}$$Then $$\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+1}{2}}+\sqrt{\frac{5-1}{2}}=\sqrt{3}+\sqrt{2}$$Then $$\sqrt{(1+\sqrt{2}+\sqrt{3})^2}-(\sqrt{2}+\sqrt{3})=1+\sqrt{2}+\sqrt{3}-\sqrt{2}-\sqrt{3}=1$$

Critiques of the method:

Spoiler

It is "easy to see" after the fact of
already having a guess of the square of the sum
of those particular three terms that will equal
the expansion you have on the left-hand side.

There is not a natural motivation for it, as
compared to setting up a scenario such as
$\displaystyle (a + b\sqrt{2} + c\sqrt{3})^2, \ \$ and solving for integers
a, b, and c.


You relied on formulas for the radical expressions
that are neither well-known nor are
readily accessible. But also, the formulas
do the work of certain steps that you should be showing yourself.


[/spoilet]

 

[Steven G]

@BigBeachBanana
We get a day off?

 

[BigBeachBanana]

@BigBeachBanana
We get a day off?

Ya'll are making it look too easy, so I needed some time to find hard ones.
Here you go.

Algebra Problem of The Day-9

The operation * is defined as follows:x*y=\frac{x+y}{1+xy} What is the value of the following expression? \{[(2*3)*4]...\}*2021=?

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