Algebra Problem of The Day-8

[BigBeachBanana]

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

 

[red12dog34]

Spoiler

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

 

[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 [math]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[/math]and [math]\frac{1}{\sqrt{5-2\sqrt{6}}}=\sqrt{5+2\sqrt{6}}=\sqrt{5+\sqrt{24}}[/math]We use the formula [math]\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}, \ \text{where} \ C=\sqrt{A^2-B}[/math]Then [math]\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+1}{2}}+\sqrt{\frac{5-1}{2}}=\sqrt{3}+\sqrt{2}[/math]Then [math]\sqrt{(1+\sqrt{2}+\sqrt{3})^2}-(\sqrt{2}+\sqrt{3})=1+\sqrt{2}+\sqrt{3}-\sqrt{2}-\sqrt{3}=1[/math]

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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