• Jun 5th 2006, 03:09 PM
kbryant05
I have only about fifty more to do at the moment and am struggling on a few more.
First problem not sure I really get this type of one
Use rational exponents to simplify each expression.

8
√x^4

Second is to simplify the indicated operation.

3 3
√5x^2 * √10y

√x+9 - √x-7=2

Fourth is:perform indicated operation. Write in the form of a + bi
(5-3i) - (6-9i)
would I do the foil on this where 30-18i+18i okay no that ones is wrong.
• Jun 5th 2006, 03:21 PM
ThePerfectHacker
Quote:

Originally Posted by kbryant05

8
√x^4

Okay,
What you have in terms of rational exponents is,
$(x^4)^{1/8}=x^{4(1/8)}=x^{1/2}=\sqrt{x}$
Quote:

Originally Posted by kbryant05
3 3
√5x^2 * √10y

Do you want me to express it as rational exponent?
Then,
$\sqrt[3]{5x^2}=(5x^2)^{1/3}$ and,
$\sqrt[3]{10y}=(10y)^{1/3}$
Thus,
$(5x^2)^{1/3}\cdot (10y)^{1/3}=[(5x^2)(10y)]^{1/3}=(50x^2y)^{1/3}$

Quote:

Originally Posted by kbryant05
√x+9 - √x-7=2

Do you mean,
$\sqrt{x+9}-\sqrt{x-7}=2$
Or,
$\sqrt{x}+9-\sqrt{x}-7=2$
If it is case #2 then it is always true because the radicals cancel and you are left with $9-7=2$ which is always true for any non-negative real number. Thus I presume it is the first one?

Quote:

Originally Posted by kbryant05
Fourth is:perform indicated operation. Write in the form of a + bi
(5-3i) - (6-9i)

Open parantheses first and watch those signs :eek: ,
$5-3i-6+9i$
Add the reals, $5-6=-1$
Add the imaginaries, $-3i+9i=6i$
Thus,
$-1+6i$
• Jun 5th 2006, 03:42 PM
kbryant05
Okay
Yes on the second one both ways if you can show me,and on the third the first part like you have it is correct not the second part.
• Jun 5th 2006, 03:46 PM
kbryant05
Question on my second problem
The problem is

3 3
√5x^2 * √ 10y

The book is saying 3
√ 50x^2y as an answer but I see you have 1/3 in there. Where does that come in? I think I am more confused??
• Jun 5th 2006, 04:32 PM
Quick
The 1/3
I'm back.

Quote:

√ 50x^2y as an answer but I see you have 1/3 in there. Where does that come in? I think I am more confused??
the 1/3 shows $\sqrt[3]{x}$ and here's why:

$x^1 \cdot x^1 = x^{1+1} = x^2$
therefore,
$x^{1/2} \cdot x^{1/2} = x^{1/2+1/2} = x^1$
notice that the same answer works for $\sqrt {x}$
$\sqrt {x} \sqrt {x} = \sqrt {x\cdot x} = \sqrt {x^2} = x$
so $x^{1/2} = \sqrt{x}$
Now we look at 1/3
$x^{1/3} \cdot x^{1/3} \cdot x^{1/3} = x^{(1/3+1/3+1/3)} = x^1$
and the same answer comes around with $\sqrt[3]{x}$
$\sqrt[3]{x}\sqrt[3]{x}\sqrt[3]{x} = \sqrt[3]{x\cdot x \cdot x} = \sqrt[3]{x^3}=x^1$
so $x^{1/3} = \sqrt[3]{x}$

$(50x^2y)^{1/3} = \sqrt[3]{50x^2y}$
• Jun 5th 2006, 04:54 PM
kbryant05
Got it
Good job Quick!!
Thanks for explaining that. You know the answers are great that they give us in the books but getting to that answer is more than a challenge sometimes for me.
thanks,
K
• Jun 5th 2006, 06:37 PM
ThePerfectHacker
Quote:

Originally Posted by kbryant05
Yes on the second one both ways if you can show me,and on the third the first part like you have it is correct not the second part.

The second case to question 2 has been explained. The solution is any non-negative number.

The first case:
$\sqrt{x+9}-\sqrt{x-7}=2$
Thus,
$\sqrt{x+9}=2+\sqrt{x-7}$
Square both sides,
$(\sqrt{x+9})^2=(2+\sqrt{x-7})^2$
Thus, use FOIL
$x+9=4+4\sqrt{x-7}+(\sqrt{x-7})^2$
Thus,
$x+9=4+4\sqrt{x-7}+x-7$
Subtract 'x' from both sides and combine numbers,
$12=4\sqrt{x-7}$
Divide by 4,
$3=\sqrt{x-7}$
Square both sides,
$9=x-7$
Thus,
$x=16$
Check this solution to see that it works!
• Jun 6th 2006, 06:30 AM
kbryant05
Genius
Finally understanding these wonderful radicals with real and imaginaries. How fun! Some of the problems start to seem so long and tedious. I think it's easy to get lost into the problems. You guys are great at making it look so simple. Thumbs up to both of you! Thanks for the quick responses also!! :)