Author Topic: Stuck Problems (Read 1633 times) Tweet Share

Christiano · « **on:** February 03, 2010, 09:00:12 pm »

Equations I got stuck on my methods pre-test. Must rectify!!

$3-(x-2)^2$

$64y^4-16y^2+1$

$(3p-5)^2-(p+2)^2$

$a^3-3a^2+a-3$

the.watchman · « **Reply #1 on:** February 03, 2010, 10:16:23 pm »

You mean solving for equaling 0?

moekamo · « **Reply #2 on:** February 04, 2010, 12:28:18 am »

Quote from: Christiano on February 03, 2010, 09:00:12 pm

Equations I got stuck on my methods pre-test. Must rectify!!

$3-(x-2)^2$

$64y^4-16y^2+1$

$(3p-5)^2-(p+2)^2$

$a^3-3a^2+a-3$

Assuming all equations are equaling 0:

1. $0=3-(x-2)^2$
$x-2=\pm\sqrt{3}$
$x=2\pm\sqrt{3}$

2. let $u=y^2\Rightarrow 0=64u^2-16u+1$
$0=(8u-1)^2$
$u=\frac{1}{8}$
$y=\pm \frac{1}{\sqrt{8}}$
$y=\pm \frac{1}{2\sqrt{2}}$

3. $0=(3p-5)^2-(p+2)^2$
using DOPS
$0=(3p-5-(p+2))(3p-5+(p+2))=(2p-7)(4p-3)$
$p=\frac{7}{2},\frac{3}{4}$

4. $0=a^3-3a^2+a-3$
Using grouping
$0=a^2(a-3)+a-3$
$0=(a^2+1)(a-3)$
$a=3$ only

EDIT: changed x to y in 2 :S

the.watchman · « **Reply #3 on:** February 04, 2010, 07:19:13 am »

Isn't question 2 solving for y?
Just something I noticed

Doesn't really matter

moekamo · « **Reply #4 on:** February 04, 2010, 09:22:47 am »

Quote from: the.watchman on February 04, 2010, 07:19:13 am

Isn't question 2 solving for y?
Just something I noticed
Doesn't really matter

haha yep, you found my delibrate mistake!

ill fix it thanks watchman

Mao · « **Reply #5 on:** February 04, 2010, 09:58:55 am »

$\frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$

Rationalizing the denominator is important for MM.

Christiano · « **Reply #6 on:** February 04, 2010, 07:01:12 pm »

Thanks peeps, sorry I forgot to mention it was factorizing the equation, as well as equaling to zero and 1. The only problem I had was factoring the equations.

m@tty · « **Reply #7 on:** February 04, 2010, 07:10:16 pm »

Quote from: Mao on February 04, 2010, 09:58:55 am

$\frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$

Rationalizing the denominator is important for MM.

Less so for Specialist?

the.watchman · « **Reply #8 on:** February 04, 2010, 07:15:21 pm »

I see moekamo didn't factorise the first one, so I guess I'll do that:

$3-(x-2)^2$

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

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

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

Christiano · « **Reply #9 on:** February 04, 2010, 08:26:42 pm »

Man, i totally forgot I could use surds.. thanks!

the.watchman · « **Reply #10 on:** February 04, 2010, 08:28:21 pm »

Quote from: Christiano on February 04, 2010, 08:26:42 pm

Man, i totally forgot I could use surds.. thanks!

No probs, it's easy to miss

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Author Topic: Stuck Problems (Read 1633 times) Tweet Share

Christiano

Stuck Problems

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