Author Topic: brightsky's Noob-tastic Question Thread (Read 15124 times) Tweet Share

pi · « **Reply #30 on:** March 13, 2011, 03:23:09 pm »

Quote from: brightsky on March 13, 2011, 02:40:59 pm

But if you take the $\sqrt{x+3}$ up into the denominator, i.e.

$\frac{\sqrt{x+3}\sqrt{2x-1} + x}{\sqrt{2x-1}}$ , letting $(x+3)(2x-1) \ge 0$ gives you different solutions...am I missing something?

Then the domain of the numerator still has to intersect with the domain of the denominator, cancelling one of the solutions of the numerator

brightsky · « **Reply #31 on:** March 13, 2011, 03:26:38 pm »

Hmmm...so:

$2x-1 > 0$ ....[1]

$(x+3)(2x-1) \ge 0$ ....[2]

From [1], $x > \frac{1}{2}$

From [2], $x \le -3$ , $x \ge \frac{1}{2}$

Solving simultaneously, we only get $x > \frac{1}{2}$

pi · « **Reply #32 on:** March 13, 2011, 03:27:40 pm »

Quote from: brightsky on March 13, 2011, 03:26:38 pm

Hmmm...so:

$2x-1 > 0$ ....[1]

$(x+3)(2x-1) \ge 0$ ....[2]

From [1], $x > \frac{1}{2}$

From [2], $x \le -3$ , $x \ge \frac{1}{2}$

Solving simultaneously, we only get $x > \frac{1}{2}$

Yes, I think thats how to do it as a combined fraction

Remember for all sum/difference/product functions:
$Domf \ \cap \ Domg$

xZero · « **Reply #33 on:** March 13, 2011, 03:30:32 pm »

Quote from: brightsky on March 13, 2011, 03:26:38 pm

Hmmm...so:

$2x-1 > 0$ ....[1]

$(x+3)(2x-1) \ge 0$ ....[2]

From [1], $x > \frac{1}{2}$

From [2], $x \le -3$ , $x \ge \frac{1}{2}$

Solving simultaneously, we only get $x > \frac{1}{2}$

thats basically what i did before oO

brightsky · « **Reply #34 on:** March 13, 2011, 03:31:56 pm »

Quote from: Rohitpi on March 13, 2011, 03:27:40 pm

Quote from: brightsky on March 13, 2011, 03:26:38 pm
Hmmm...so:

$2x-1 > 0$ ....[1]

$(x+3)(2x-1) \ge 0$ ....[2]

From [1], $x > \frac{1}{2}$

From [2], $x \le -3$ , $x \ge \frac{1}{2}$

Solving simultaneously, we only get $x > \frac{1}{2}$

Yes, I think thats how to do it as a combined fraction

Remember for all sum/difference/product functions:
$Domf \ \cap \ Domg$

Omg..haha I just realised x >= -3 was redundant from my first post. aiya >< Thanks Rohitpi and ZeroX!

xZero · « **Reply #35 on:** March 13, 2011, 03:34:08 pm »

Quote from: Rohitpi on March 13, 2011, 03:19:58 pm

Quote from: xZero on March 13, 2011, 03:14:30 pm
really?
$(x+3)(2x-1)\ge 0$

$(x+3)\ge 0$ and $(2x-1)\ge 0$

$x\ge -3$ and $x\ge \frac{1}{2}$

I thought you were not able to solve quadratic inequations like that. I though a mental sketch graph was necessary... Which would make $x \le -3$ and $x \ge \frac{1}{2}$ correct.

EDIT: Wolfram Alpha agrees too

well you're correct if the denominator doesn't exist, however in this question $x\le -3$ was not valid as both brackets must be positive, thats why i skipped that step

pi · « **Reply #36 on:** March 13, 2011, 03:36:56 pm »

Oh, sorry for doubting it then!

VCE247 · « **Reply #37 on:** March 14, 2011, 07:35:21 pm »

Brightsky is this methods 3+4 O_O

nacho · « **Reply #38 on:** March 14, 2011, 07:43:15 pm »

Quote from: VCE247 on March 14, 2011, 07:35:21 pm

Brightsky is this methods 3+4 O_O

yea,
brightsky is Derreck Ha's successor.

m@tty · « **Reply #39 on:** March 14, 2011, 08:06:13 pm »

Quote from: nacho on March 14, 2011, 07:43:15 pm

Quote from: VCE247 on March 14, 2011, 07:35:21 pm
Brightsky is this methods 3+4 O_O
yea,
brightsky is Derreck Ha's successor.

I believe it goes like this:

Derrick Ha < James Lu < brightsky

haha

pi · « **Reply #40 on:** March 14, 2011, 08:09:30 pm »

Quote from: m@tty on March 14, 2011, 08:06:13 pm

I believe it goes like this:

Derrick Ha < James Lu < brightsky haha

+1, I think thats about right

brightsky · « **Reply #41 on:** March 14, 2011, 08:38:22 pm »

lolwut?! I think putting thushan in the place of that guy called "brightsky" would be more applicable. This is the height of overestimation.

vea · « **Reply #42 on:** March 14, 2011, 08:43:38 pm »

Quote from: brightsky on March 14, 2011, 08:38:22 pm

lolwut?! I think putting thushan in the place of that guy called "brightsky" would be more applicable. This is the height of overestimation.

True, brightsky completes his VCE in 2013...
So: Derrick Ha<James Lu<thushan<brightsky

iNerd · « **Reply #43 on:** March 14, 2011, 08:44:30 pm »

Quote from: vea on March 14, 2011, 08:43:38 pm

Quote from: brightsky on March 14, 2011, 08:38:22 pm
lolwut?! I think putting thushan in the place of that guy called "brightsky" would be more applicable. This is the height of overestimation.

True, brightsky completes his VCE in 2013...
So: Derrick Ha<James Lu<thushan<brightsky

.

Sorry brightsky but that was extremely smart.

brightsky · « **Reply #44 on:** March 15, 2011, 08:31:12 pm »

Can someone clarify, does R^+ / R^- include 0 or not? I remember reading something about this on VN but I can't find it. :/

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