Solution to Mr Alk...'s Questions (1 Viewer)

Drongoski · Apr 5, 2011

Q1) Factorise:

a) $a^2c - b^2c -abc^2 + ab$

b) $a^2 - b^2 + 2a + 1$

Q2) Find x & y if :

$\text { } x - \sqrt{3} y = \sqrt {\frac {2-\sqrt{3}}{ 2+\sqrt{3}}}$

Q3) Simplify:

$\frac {x^{\frac {3}{2}}+y^{\frac{3}{2}}}{x^{\frac{1}{2}} - y^{\frac {1}{2}}}}$

Solutions

a)

$= (a^2c - abc^2) + (ab - b^2c)$

$= ac(a-bc) + b(a-bc)$

$= (ac + b)(a - bc)$

b)

$= a^2 +2ab+1 - b^2$

$= (a+1)^2 - b^2$

$= (a+1 + b)(a +1 - b)$

Q2)

$x - \sqrt {3} y = \sqrt{\frac {(2-\sqrt {3}) (2-\sqrt{3})}{(2+ \sqrt{3})(2-\sqrt{3}))}}$

$= \sqrt {\frac {(2-\sqrt{3})^2}{(2^2 - (\sqrt {3})^2)}}$

$= 2 - \sqrt {3}$

Equating corresponding parts:

$x = 2 \text { and } y = 1$

Q3)

This is simply doing:

$\frac {A^3 + B^3}{A+B} = \frac {(A+B)(A^2 - AB + B^2)}{A+B}$

$\therefore \text {ans }= x - \sqrt{xy} + y$

ohexploitable · Apr 5, 2011

Who is Mr. Alk? Are these meant to be hard questions?

Drongoski · Apr 5, 2011

He PM'd me for private solutions. But I must respect his privacy.

Are they meant to be hard? Don't know. What do u think? I suspect the average 3U student would struggle with a couple of them at first.

I also like the way I compete against myself. My free-service arm is doing a better business. Soon I'll have to queue up at Centrelink.

Drongoski · Apr 5, 2011

ohexploitable said:
Who is Mr. Alk?

Hey you crazy or somethin. I can't afford to disclose my client's identity. My competitors will be after his business!

SpiralFlex · Apr 5, 2011

Drongoski said:
Hey you crazy or somethin. I can't afford to disclose my client's identity. My competitors will be after his business!

I answered this question for him also. He told me to use paper, he did not like Latex. ^_^

Drongoski · Apr 5, 2011

SpiralFlex said:
I answered this question for him also. He told me to use paper, he did not like Latex. ^_^

We are deadly competitors now. How much did he pay you?

SpiralFlex · Apr 5, 2011

Lol, I did it for PHREEE! ;-)

Drongoski · Apr 5, 2011

SpiralFlex said:
Lol, I did it for PHREEE! ;-)

To compete successfully against you, I may have to pay him for the privilege of doing his questions from now on. It's a life-&-death struggle for survival in this business.

SpiralFlex · Apr 5, 2011

Drongoski said:
To compete successfully against you, I may have to pay him for the privilege of doing his questions from now on.

Alkanes · Apr 6, 2011

Thank You guys for both of you for offering help. I actually asked you Drongoski first but took quite a while to respond and i was in need of a rush so i got help from the next genius Spiral =)

Drongoski · Apr 6, 2011

Don't worry. I didn't see your PM earlier and for some reason I could not open the image. Besides I was away making a meagre living giving private tuition.

Solution to Mr Alk...'s Questions (1 Viewer)

Drongoski

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ohexploitable

hey

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