• September 23rd 2012, 02:19 PM
Benja303
Hi, I am new to the forum. I have been stuck on a few math problems involving exponents for a while now and i don't know where else to turn.

I have attached a pdf file with my work and the questions i need help with are as follows. * These are not going to be marked and we were given most of the answers, so please explain to me and don't just give me the answer

as you can see all the answers are already given so i'm not trying to cheat.. or be lazy and get people to work for me.. I benefit in no way whatsoever.

section a

2.a) Okay the answer for this one is supposed to be 3 √1/32
But i don't understand how to arrive at this. I really don't even know where to start because i don't understand my teacher (currently working on switching teachers)

2.b) same as 2.a idk..

The rest i get

Then to section B
9.) i)
ii)

Please explain these. There are more i don't understand but i'm hoping by you explaining these to me i will be able to complete the rest.
• September 23rd 2012, 03:08 PM
Siron
2.a Multiply each side with $x^{\frac{1}{2}}$
2.b
$x^{\frac{1}{2}} +\frac{1}{2}x^{\frac{-1}{2}}(x-9)=0$
$\Leftrightarrow x^{\frac{1}{2}}+\frac{1}{2}x^{\frac{-1}{2}}x - \frac{9}{2}x^{\frac{-1}{2}}=0$
$\Leftrightarrow x^{\frac{1}{2}}+\frac{1}{2}x^{\frac{1}{2}}-\frac{9}{2}x^{\frac{-1}{2}}=0$
Same here, mulitply each side with $x^{\frac{1}{2}}$.
• September 23rd 2012, 03:15 PM
pickslides
$\displaystyle x^{\frac{-3}{2}}-\frac{1}{9}x^{\frac{-1}{2}} = 0$

Make $\displaystyle x^{\frac{-1}{2}} = u$ then you have

$\displaystyle u^3-\frac{1}{9}u = 0$ solve this cubic by factoring out $u$ then applying the difference of two squares.

You should get $\displaystyle u = 0, \frac{-1}{3}, \frac{1}{3} \implies x = \frac{1}{\sqrt{0}}, \frac{1}{\sqrt{\frac{-1}{3}}}, \frac{1}{\sqrt{\frac{1}{3}}}
$

Throw out the first two answers for reasons linked with the domain of the real numbers.
• September 23rd 2012, 03:24 PM
skeeter
2. (a) I do not agree with your given solution. The expression will factor ...

$x^{-3/2} \left(1 - \frac{x}{9}\right) = 0$

$x = 9$ is the only solution.

2. (b) this expression will also factor ...

$x^{-1/2} \left(x + \frac{x-9}{2} \right) = 0$

$x = 3$ is the only solution

problem 9 follows the same procedure as problem 2 ...

9. (i) factor out $x^{-1/3}$

(ii) factor out $x(x+5)^{-1/2}$
• September 23rd 2012, 07:29 PM
bjhopper
For 2a the exponents are positive not negative as i view them with a magnifying glassand my answer is 1/9
• September 24th 2012, 04:07 AM
skeeter