What about these types of radicals?

**kbryant05** · June 5th 2006, 03:09 PM

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

Third is:solving this radical equation:
√x+9 - √x-7=2

Fourth is

erform 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.

**ThePerfectHacker** · June 5th 2006, 03:21 PM

Originally Posted by kbryant05

8
√x^4

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

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,
$(5x^2)^{1/3}\cdot (10y)^{1/3}=[(5x^2)(10y)]^{1/3}=(50x^2y)^{1/3}$

Originally Posted by kbryant05

Third is:solving this radical equation:
√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?

Originally Posted by kbryant05

Fourth is

erform indicated operation. Write in the form of a + bi
(5-3i) - (6-9i)

Open parantheses first and watch those signs

,
$5-3i-6+9i$
Add the reals, $5-6=-1$
Add the imaginaries, $-3i+9i=6i$
Thus,
$-1+6i$

**kbryant05** · June 5th 2006, 03:42 PM

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.

**kbryant05** · June 5th 2006, 03:46 PM

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??

**Quick** · June 5th 2006, 04:32 PM

I'm back.

√ 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 $x^{1/3} = \sqrt[3]{x}$

which makes Hacker's answer,
$(50x^2y)^{1/3} = \sqrt[3]{50x^2y}$
which is the answer in your book.

**kbryant05** · June 5th 2006, 04:54 PM

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

**ThePerfectHacker** · June 5th 2006, 06:37 PM

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!

**kbryant05** · June 6th 2006, 06:30 AM

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!!

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What about these types of radicals?

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Question on my second problem

The 1/3

Got it

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