## Wednesday, February 3, 2016

### Analysis Quiz 19: Multiple-Choice Test (Improve Analytical Skill)

Question 1: If you're asked to simplify [MATH]\left(1+\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\right)^2[/MATH], do you think by turning the $1$ as [MATH]\left(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\right)[/MATH] is feasible in order to simplify the expression?

A. Yes.
B. No.

Yes, of course, since that could be the good strategy to start working on the problem.

Question 2: If you answered yes to question 1, then what would you expect how the expression [MATH]\frac{(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)+(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}[/MATH] would look like?

A. [MATH]\frac{a(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)}{(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)}[/MATH], with [MATH]a[/MATH] an irrational number.

B. [MATH]\frac{a(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)}{(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)}[/MATH], with [MATH]a[/MATH], with [MATH]a\in \Bbb{N}[/MATH].

C. [MATH]\frac{a(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)}{(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)}[/MATH], with [MATH]a[/MATH], with [MATH]a\gt 0[/MATH].

D. [MATH]\frac{a(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)}{(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)}[/MATH], with [MATH]a[/MATH], with [MATH]a\lt 0[/MATH].

We can rule out the option D at once since we know the given target expression [MATH]1+\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}[/MATH] must be a positive term.

The chance that $a$ will be an irrational number or a natural number is 50-50, and we wouldn't know it unless we work on it. And the option C that speculates $a\gt 0$ must be correct.

Therefore, we have to check the first three options as the answer for this question.

Question 3: How are you going to simplify [MATH](\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)+(\sqrt{2}+\sqrt{3}+\sqrt{4})[/MATH]?

A. First collect [MATH]\sqrt{2}+\sqrt{3}+\sqrt{2}+\sqrt{3}[/MATH] and then look for the common factor of the whole expression to see if we can further factorize it.

B. After collecting $\sqrt{2}+\sqrt{3}+\sqrt{2}+\sqrt{3}$as $2(\sqrt{2}+\sqrt{3})$, and turning the sum into $2(\sqrt{2}+\sqrt{3})+\sqrt{4}+\sqrt{6}+\sqrt{8}+4$, there is nothing left to simplify.

We will of course first collect [MATH]\sqrt{2}+\sqrt{3}+\sqrt{2}+\sqrt{3}[/MATH] soo that we get:

[MATH](\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)+(\sqrt{2}+\sqrt{3}+\sqrt{4})[/MATH]

[MATH]=2(\sqrt{2}+\sqrt{3})+\sqrt{6}+\sqrt{8}+4+\sqrt{4}[/MATH]

[MATH]=2\sqrt{2}+2\sqrt{3}+\sqrt{2}\sqrt{3}+\sqrt{4}\sqrt{2}+4+\sqrt{4}[/MATH](*)

It's tempting now to jump into conclusion that the expression (*) above couldn't be simplified further as there is no common factor. But wait a minute, note that $\sqrt{2}$ exists in

[MATH]=2\color{red}\sqrt{2}\color{black}+2\sqrt{3}+\color{red}\sqrt{2}\color{black}\sqrt{3}+\sqrt{4}\color{red}\sqrt{2}\color{black}+4+\sqrt{4}[/MATH]

for those red terms, so what we want to go is we want to turn $2$ as $\sqrt{2}\sqrt{2}$, that yields:

[MATH]2\color{red}\sqrt{2}\color{black}+2\sqrt{3}+\color{red}\sqrt{2}\color{black}\sqrt{3}+\sqrt{4}\color{red}\sqrt{2}\color{black}+4+\sqrt{4}[/MATH]

[MATH]=2\color{red}\sqrt{2}\color{black}+\color{red}\sqrt{2}\sqrt{2}\color{black}\sqrt{3}+\color{red}\sqrt{2}\color{black}\sqrt{3}+\sqrt{4}\color{red}\sqrt{2}\color{black}+\sqrt{16}+\sqrt{2}\sqrt{2}[/MATH]

[MATH]=2\color{red}\sqrt{2}\color{black}+\color{red}\sqrt{2}\sqrt{2}\color{black}\sqrt{3}+\color{red}\sqrt{2}\color{black}\sqrt{3}+\sqrt{4}\color{red}\sqrt{2}\color{black}+\color{red}\sqrt{2}\color{black}\sqrt{8}+\color{red}\sqrt{2}\color{black}\sqrt{2}[/MATH]

[MATH]=\color{red}\sqrt{2}\color{yellow}\bbox[5px,purple]{(2+\sqrt{2}\sqrt{3}+\sqrt{3}+\sqrt{4}+\sqrt{8}+\sqrt{2})}[/MATH]

Now, what we ought to bear in mind is, the second factor above, i.e. [MATH]\color{yellow}\bbox[5px,purple]{(2+\sqrt{2}\sqrt{3}+\sqrt{3}+\sqrt{4}+\sqrt{8}+\sqrt{2})}[/MATH] most probably could be rewritten to take the form

[MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4[/MATH]

Let's see:

[MATH]\color{yellow}\bbox[5px,purple]{(2+\sqrt{2}\sqrt{3}+\sqrt{3}+\sqrt{4}+\sqrt{8}+\sqrt{2})}[/MATH]

[MATH]=2+\sqrt{2\cdot 3}+\sqrt{3}+\sqrt{4}+\sqrt{8}+\sqrt{2}[/MATH]

[MATH]=2+\sqrt{6}+\sqrt{3}+2+\sqrt{8}+\sqrt{2}[/MATH]

[MATH]=4+\sqrt{6}+\sqrt{3}+\sqrt{8}+\sqrt{2}[/MATH]

[MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4[/MATH]

[MATH]\therefore (\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)+(\sqrt{2}+\sqrt{3}+\sqrt{4})=\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4)[/MATH]

Question 4: Do you think you could simplify the fraction [MATH]\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}[/MATH]?

A. Yes, by rationalizing the denominator [MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4[/MATH].
B. Yes, by not rationalizing the denominator [MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4[/MATH].

Yes, we could definitely simplify the fraction [MATH]\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}[/MATH], and given the denominator consists of more than 2 terms, it's a bit of a stretch to try it from the perspective of rationalizing it.

Plus, we might want to suspect if the denominator could be factored nicely where one of the factors is equivalent to the expression of the numerator.

Question 5: How you would begin to rationalize the denominator [MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4[/MATH]?

A. By multiplying it with [MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}-4[/MATH].
B. By multiplying it with [MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}-\sqrt{8}-4[/MATH].
C. By multiplying it with [MATH]\sqrt{2}+\sqrt{3}-\sqrt{6}-\sqrt{8}-4[/MATH].
D. By multiplying it with [MATH]\sqrt{2}-\sqrt{3}-\sqrt{6}-\sqrt{8}-4[/MATH].
E. Neither of the above.

As we've already mentioned in the discussion for question 4, we suspected the denominator could be factored into $(\sqrt{2}+\sqrt{3}+\sqrt{4})(P(x))$, therefore, we don't want to attempt it from rationalizing the denominator approach.

The answer for this question is hence E.

Question 6: If [MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4=P(x)(\sqrt{2}+\sqrt{3}+\sqrt{4})[/MATH], what would you do to find $P(x)$?

A. By working with [MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4[/MATH] and try to factor it.
B. By doing the long division.

The simplest way to determine if [MATH]\sqrt{2}+\sqrt{3}+\sqrt{4}[/MATH] is a factor of [MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4[/MATH] is through the long division method:

[MATH]\begin{array}{r}1+\sqrt{2}\hspace{33px}\\\sqrt{2}+\sqrt{3}+\sqrt{4}\enclose{longdiv}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4} \\ -\underline{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)} \hspace{60px} \\ 2+\sqrt{6}+\sqrt{8}\hspace{33px}\\-\underline{\left(2+\sqrt{6}+\sqrt{8}\right)} \hspace{20px} \end{array}[/MATH]

Therefore, we can safely say [MATH]\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4=(1+\sqrt{2})(\sqrt{2}+\sqrt{3}+\sqrt{4})[/MATH].

Question 7: What is the simplified answer for [MATH]\left(1+\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\right)^2[/MATH]?
A. $1+\sqrt{2}$
B. $\sqrt{2}$
C. $2$

[MATH]\begin{align*}\left(1+\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\right)^2&=\left(1+\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{(1+\sqrt{2})(\sqrt{2}+\sqrt{3}+\sqrt{4})}\right)^2\\&=\left(1+\frac{\cancel{\sqrt{2}+\sqrt{3}+\sqrt{4}}}{(1+\sqrt{2})(\cancel{\sqrt{2}+\sqrt{3}+\sqrt{4}})}\right)^2\\&=\left(1+\frac{1}{1+\sqrt{2}}\right)^2\\&=\left(1+\frac{(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}\right)^2\\&=\left(1+\frac{(1-\sqrt{2})}{-1}\right)^2\\&=(1-1+\sqrt{2})^2\\&=2\end{align*}[/MATH]

Question 8: Can you think of any other method(s) that is/are different than the previously discussed two methods(turning $1$ as [MATH]\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}[/MATH] or to simplify  [MATH]\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}[/MATH]) to simplify the given square?
No, those two methods (first by turning $1$ as [MATH]\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}[/MATH] and proceed from there and the second by simplifying the fraction  [MATH]\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}[/MATH]) are the only methods one could use to simplify the expression.