# Thread: 3 quite good problems (mostly algebra)

1. ## 3 quite good problems (mostly algebra)

Hello everybody!

I hope you feel for some algebra!

Problem 1 (I have "solved" this one, "solution" below)

$\displaystyle \frac {4}{\sqrt {5} - 3} + 3 + \ \sqrt5$

Spoiler! - UPDATED
This is how I think

First we put $\displaystyle 3 + \sqrt5$ as a fraction

$\displaystyle \frac {4}{\sqrt {5} - 3} + \frac {3 + \sqrt5} {1}$

Then we multiply $\displaystyle \frac {3 + \sqrt5} {1}$ with $\displaystyle \sqrt {5} - 3$ to get a common denominator.

$\displaystyle \frac {4}{\sqrt {5} - 3} + \frac {(\sqrt {5} - 3)3 + \sqrt5} {\sqrt {5} - 3}$

Ok, now we are close to a more simpler way to put this problem. (Put everything in the same fraction.

$\displaystyle \frac {4 + (\sqrt {5} - 3)3 + \sqrt5} {\sqrt {5} - 3}$

$\displaystyle 4 + (\sqrt {5} - 3)(\sqrt {5} + 3)$ is zero. we are left with

$\displaystyle \frac {0} {\sqrt{5} - 3} = 0$

End Spoiler

A second opinion on this problem and it's "solution" is always welcome.

Alright, problem 2. Now we got to have a clear mind and a sharp eye to solve this one...

$\displaystyle \frac {a}{ \sqrt{ab} + a} + \frac {b} {\sqrt{ab} - b} - \frac {a}{a-b}$

Solved!

First we work some on the first fraction, to make it more easy to work with. (I'll go more into detail later)

Problem 3

$\displaystyle \displaystyle {\frac{(9 \cdot 16^{n - 1} + 16^n)^2} {(4^{n - 1} + 4^{n - 2})^4}}$

Solved!

2. Originally Posted by λιεҗąиđ€ŗ
Problem 1 (I have "solved" this one, "solution" below)

$\displaystyle \frac {4}{\sqrt {5} - 3} + 3 + \ \sqrt5$
$\displaystyle \frac4{\sqrt5-3}+3+\sqrt5=\frac4{\sqrt5-3}+\left(\sqrt5+3\right)\cdot\frac{\sqrt5-3}{\sqrt5-3}$

Does that make sense?

--

I'm gonna eat now, but probably Jhevon's gonna help ya with the second one

3. Originally Posted by λιεҗąиđ€ŗ
Alright, problem 2. Now we got to have a clear mind and a sharp eye to solve this one...

$\displaystyle \frac {a}{ \sqrt{ab} + a} + \frac {b} {\sqrt{ab} - b} - \frac {a}{a-b}$

Okay, so let's take this

$\displaystyle \frac a{\sqrt{ab}+a}=\frac{\sqrt a\sqrt a}{\sqrt a\sqrt b+\sqrt a\sqrt a}=\frac{\sqrt a}{\sqrt b+\sqrt a}$ (I assume you know how to kill this)

Do the same with $\displaystyle \frac b{\sqrt{ab}-b}$

4. Originally Posted by λιεҗąиđ€ŗ
$\displaystyle \frac {a}{ \sqrt{ab} + a} + \frac {b} {\sqrt{ab} - b} - \frac {a}{a-b}$
Alternatively, multiply by the conjugate straight away

(Looks right to me, but my calculator doesn't say it's true)?

$\displaystyle \frac {a}{ \sqrt{ab} + a}=\frac {a}{ \sqrt{ab} + a} \times \frac{\sqrt{ab}-a}{\sqrt{ab}-a}=\frac{a(\sqrt{ab}-a)}{ab-a^2}=\frac{\sqrt{ab}-a}{b-a}$ (and to give it a common demoninator) $\displaystyle =-\frac{\sqrt{ab}-a}{a-b}$

5. Originally Posted by Krizalid
$\displaystyle \frac4{\sqrt5-3}+3+\sqrt5=\frac4{\sqrt5-3}+\left(\sqrt5+3\right)\cdot\frac{\sqrt5-3}{\sqrt5-3}$

Does that make sense?

--

I'm gonna eat now, but probably Jhevon's gonna help ya with the second one
$\displaystyle \left(\sqrt5+3\right)\cdot\frac{\sqrt5-3}{\sqrt5-3}$

It does make sense, one gets a common denominator with this operation, correct?

And the problem would look like this then:

$\displaystyle \frac {4}{\sqrt {5} - 3} + \frac {(\sqrt {5} - 3)3 + \sqrt5} {\sqrt {5} - 3}$

P.S.
Hola de Suecia

Espero que su desayuno fuera bueno

6. Originally Posted by DivideBy0
Alternatively, multiply by the conjugate straight away

(Looks right to me, but my calculator doesn't say it's true)?

$\displaystyle \frac {a}{ \sqrt{ab} + a}=\frac {a}{ \sqrt{ab} + a} \times \frac{\sqrt{ab}-a}{\sqrt{ab}-a}=\frac{a(\sqrt{ab}-a)}{ab-a^2}=\frac{\sqrt{ab}-a}{b-a}$ (and to give it a common demoninator) $\displaystyle =-\frac{\sqrt{ab}-a}{a-b}$
Yes problem 2... I also try for a cd (common denominator). I suppose this would be something like a merged fraction:

$\displaystyle \frac {a(\sqrt{ab} - b)(a - b)+b(\sqrt {ab} + a)(a - b) - a (\sqrt {ab} + a)(\sqrt {ab} - b)} {(\sqrt{ab} + a)(\sqrt{ab} - b)(a-b)}$

Hmmm... the denominator seems fishy though, doesn't it?
Note: After a bit of thinking, I guess it is correct.

P.S.
Sorry for the double post, got carried away by all the math.

7. Originally Posted by λιεҗąиđ€ŗ

$\displaystyle \frac {a}{ \sqrt{ab} + a} + \frac {b} {\sqrt{ab} - b} - \frac {a}{a-b}$
There are 3 fractions. Since the first two have the form (x+y),(x-z) with their denominators, combine them into one fraction only so the sqrt(ab) will be cleared in the denominator.

= [a(sqrt(ab) -b) +b(sqrt(ab) +a] /[(sqrt(ab) +a)(sqrt(ab) -b)]
= [a*sqrt(ab) -ab +b*sqrt(ab) +ab] /[ab -b*sqrt(ab) +a*sqrt(ab) -ab]
= [(a+b)sqrt(ab)] /[(a-b)sqrt(ab)]
= (a+b) /(a-b)

Now combine that to the 3rd fraction,
(a+b)/(a-b) -a/(a-b)
= (a+b -a)/(a-b)

8. Originally Posted by λιεҗąиđ€ŗ

$\displaystyle \frac {4}{\sqrt {5} - 3} + 3 + \ \sqrt5$
Just combine them all into one fraction only,
common denominator is (sqrt(5) -3)

4/(sqrt(5) -3) +3 +sqrt(5)
= [4 +3(sqrt(5) -3) +sqrt(5)*(sqrt(5) -3)] /[sqrt(5) -3]
= [4 +3sqrt(5) -9 +5 -3sqrt(5)] /[sqrt(5) -3]
= 0/[sqrt(5) -3]

9. Originally Posted by ticbol
There are 3 fractions. Since the first two have the form (x+y),(x-z) with their denominators, combine them into one fraction only so the sqrt(ab) will be cleared in the denominator.

= [a(sqrt(ab) -b) +b(sqrt(ab) +a] /[(sqrt(ab) +a)(sqrt(ab) -b)]
= [a*sqrt(ab) -ab +b*sqrt(ab) +ab] /[ab -b*sqrt(ab) +a*sqrt(ab) -ab]
= [(a+b)sqrt(ab)] /[(a-b)sqrt(ab)]
= (a+b) /(a-b)

Now combine that to the 3rd fraction,
(a+b)/(a-b) -a/(a-b)
= (a+b -a)/(a-b)
I'm putting that in formula to get a better view of it.

$\displaystyle \frac{a(\sqrt(ab) -b) +b(\sqrt(ab) +a} {(\sqrt(ab) +a)(\sqrt(ab) -b)}$

$\displaystyle \frac {a*\sqrt(ab) - ab + b*\sqrt(ab) + ab} {ab - b*\sqrt(ab) +a*\sqrt(ab) -ab}$

$\displaystyle \frac {(a+b)\sqrt(ab)} {(a-b)\sqrt(ab)}$

$\displaystyle \frac{(a+b)} {(a-b)}$

Originally Posted by ticbol
Just combine them all into one fraction only,
common denominator is (sqrt(5) -3)

4/(sqrt(5) -3) +3 +sqrt(5)
= [4 +3(sqrt(5) -3) +sqrt(5)*(sqrt(5) -3)] /[sqrt(5) -3]
= [4 +3sqrt(5) -9 +5 -3sqrt(5)] /[sqrt(5) -3]
= 0/[sqrt(5) -3]
Step 1
$\displaystyle \frac {4}{(\sqrt(5) -3) + 3 +\sqrt5}$
Step 2
$\displaystyle \frac {4 +3(\sqrt(5) -3) +\sqrt(5)*(\sqrt(5) -3)} {\sqrt(5) -3}$
Step 3
$\displaystyle \frac {4 +3\sqrt(5) -9 +5 -3\sqrt(5)} {\sqrt(5) -3}$
Step 4
$\displaystyle \frac {0}{\sqrt(5) -3}$

$\displaystyle 0$

ticbol, do I understand you correctly? I disagree with you concerning step 1, 2.

10. Originally Posted by λιεҗąиđ€ŗ
Step 1
$\displaystyle \frac {4}{(\sqrt(5) -3) + 3 +\sqrt5}$
Step 2
$\displaystyle \frac {4 +3(\sqrt(5) -3) +\sqrt(5)*(\sqrt(5) -3)} {\sqrt(5) -3}$
Step 3
$\displaystyle \frac {4 +3\sqrt(5) -9 +5 -3\sqrt(5)} {\sqrt(5) -3}$
Step 4
$\displaystyle \frac {0}{\sqrt(5) -3}$

$\displaystyle 0$

ticbol, do I understand you correctly? I disagree with you concerning step 1, 2.
I'm "quoting" the second part now. What is shown are "skeletons" of LaTex. How could I explain these now?

Anyway, step 1 is not 4 over all of those.

Rather, step 1 is, [4/(sqrt(5) -3)] +[3] +[sqrt(5)]
Or, [4/(sqrt(5) -3)] +[3/1] +[sqrt(5) /1]

The +3 +sqrt(5) are not under the 4.

See again my solution.

----------------------------------

Now, suppose your step 1 is correct, how did you arrive to your step 2?

-----------------------------------------

ticbol, do I understand you correctly? I disagree with you concerning step 1, 2.

No, you do not understand my solution.
Yes, you should disagree with yourself concerning your steps 1 and 2.

11. I'm "quoting" the second part now. What is shown are "skeletons" of LaTex. How could I explain these now?

Anyway, step 1 is not 4 over all of those.

Rather, step 1 is, [4/(sqrt(5) -3)] +[3] +[sqrt(5)]
Or, [4/(sqrt(5) -3)] +[3/1] +[sqrt(5) /1]

The +3 +sqrt(5) are not under the 4.

See again my solution.
Haha do you prefer raw latex code? Please put it in formula, I can't follow your method by raw code.

Now, suppose your step 1 is correct, how did you arrive to your step 2?
I worked some more on it and here is what I have:

$\displaystyle \frac {a}{ \sqrt{ab} + a} + \frac {b} {\sqrt{ab} - b} - \frac {a}{a-b}$

Now we multiply each fraction with the two different denominators, we'll get a common denominator, $\displaystyle (\sqrt{ab} + a)(\sqrt{ab} - b)(a - b)$, with this method.

$\displaystyle \frac {a}{\sqrt{ab} + a} \times \frac {(\sqrt{ab} -b)(a - b)}{(\sqrt{ab} -b)(a - b)}$

$\displaystyle \frac {b}{\sqrt{ab} - a} \times \frac {(\sqrt{ab} + a)(a - b)}{(\sqrt{ab} + a)(a - b)}$

$\displaystyle \frac {a}{a - b} \times \frac {(\sqrt{ab} + a)(\sqrt{ab} - a)}{(\sqrt{ab} + a)(\sqrt{ab} - a)}$

Now we can easily put them together

$\displaystyle \frac {a(\sqrt{ab} - b)(a - b)}{(\sqrt{ab} + a)(\sqrt{ab} - b)(a-b)} + \frac {b(\sqrt {ab} + a)(a - b)} {(\sqrt{ab} + a)(\sqrt{ab} - b)(a-b)} - \frac {a (\sqrt {ab} + a)(\sqrt {ab} - b)}{(\sqrt{ab} + a)(\sqrt{ab} - b)(a-b)}$

$\displaystyle \frac {a(\sqrt{ab} - b)(a - b)+b(\sqrt {ab} + a)(a - b) - a (\sqrt {ab} + a)(\sqrt {ab} - b)} {(\sqrt{ab} + a)(\sqrt{ab} - b)(a-b)}$

Now some parts neutralizes each other, I'll skip the removal part because I can't picture it in Latex, (maybe if I knew how to use the horizontal line command).

The denominator is neutralized, and corresponding amount is removed from the numerator.

However, going from here gets wierd, we can remove different sets and thus get an different answer, can't we?

Example:

$\displaystyle a + b(a - b) - a(\sqrt{ab} + a)(\sqrt{ab} - b)$

Let's open up the parentises

$\displaystyle a + ba - b^2 - a(a\sqrt{ab} - b\sqrt{ab})$

Note: I am not 100% sure of the following

$\displaystyle a + ab - b^2 - a^2\sqrt{ab} + ab\sqrt{ab}$

Note: Oh my gosh! Wrong operator! Quick correction

@ Khrizalid

I have worked some on your method by dividing things up into square roots, but I am not coming anywhere with it.

$\displaystyle \frac {\sqrt{a}\sqrt{a}}{\sqrt{a}\sqrt{b}+\sqrt{a}\sqrt{ a}} + \frac {\sqrt{b}\sqrt{b}} {\sqrt{a}\sqrt{b} - \sqrt{b}\sqrt{b}} - \frac {\sqrt{a}\sqrt{a}} {\sqrt{a}\sqrt{a}-\sqrt{b}\sqrt{b}}$

You have failed to multiply sqrt(5) - 3 properly in the second term.
The second term should read (sqrt(5) - 3)(3+sqrt(5)).
This results in 5 - 9 which is -4.
The first term is +4 so you get 0 over the common denominator of sqrt(5) - 3.

In the second problem, the denominator can be reduced to (a-b)^2(sqrt(ab))

13. Originally Posted by oaksoft

You have failed to multiply sqrt(5) - 3 properly in the second term.
The second term should read (sqrt(5) - 3)(3+sqrt(5)).
This results in 5 - 9 which is -4.
The first term is +4 so you get 0 over the common denominator of sqrt(5) - 3.

In the second problem, the denominator can be reduced to (a-b)^2(sqrt(ab))
I have failed to multiply the sqrt(5) -3 properly.
Is that so?

But my failure (failing?) led to zero also? Oh.

In the second problem, the denominator can be reduced to (a-b)^2(sqrt(ab))
I did not go that route, remember?
Mine was easier.

14. Originally Posted by ticbol
I have failed to multiply the sqrt(5) -3 properly.
Is that so?

But my failure (failing?) led to zero also? Oh.

In the second problem, the denominator can be reduced to (a-b)^2(sqrt(ab))
I did not go that route, remember?
Mine was easier.
I could have probably made that clearer when I responded but we would have missed the fun of awkwardness .

15. ## Problem 1: Update

I'm back after a time off with no Internet acess

I have reviewed problem 1 and realized that it bottles down to $\displaystyle \frac { 4 + (\sqrt{5} - 3)(\sqrt{5} + 3)} { (\sqrt{5} - 3) }$

The nominator simplified is 4 + 5 - 9, which is zero.

And our final answer is zero.

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