Simply Radicals?!!??

[lovebugjay]

So, I have no clue how or where to even start...does anyone know how to simply radicals all are really know is that all exponents inside the square root sign are pos and less than , no common factor can be removed from n and all exponents...the denominator is rationalized...but what do i do to solve the problem???

 

[lovebugjay]

it might have helped if i had show you alll this in the first place sorry!!!

(1/x^3)*3√y^8

 

[jwpaine]

take it in pieces:

sqrt(y^8) = y^4 because y^4*y*4 = y*8

so 3sqrt(y^8) = 3y^4

now you have (1/x^3)(3x^4)

(1/x^3)*(3x^4)/1 = (3x^4)/(x^3)

Done.

 

[skeeter]

it might have helped if i had show you alll this in the first place sorry!!!

(1/x^3)*3√y^8 = 3y⁴/x³

sorry if I fail to see what is difficult about simplifying this expression ...

perhaps if you showed the original problem in its entirety?

 

[lovebugjay]

i guess i still just don't understand...as far as i can get is the index...of 3 I knew that much but everything else I don't get how it happend

 

[lovebugjay]

the answer according to my teacher should be (y^2/x^3) 3√y^2

 

[skeeter]

did you mean for your expression to be

\(\displaystyle \L \frac{1}{x^3} \cdot 3\sqrt{y^8}\)

or

\(\displaystyle \L \frac{1}{x^3} \cdot \sqrt[3]{y^8}\)

???

 

[lovebugjay]

the last one

 

[lovebugjay]

like my next problem in

5√y^7 but the five needs to be above the √ like the three was i just don't know how to show that....sorry I am not very good at this stuff as you can see!!

It asked me for my index and I stated it was five so I got 1/2 a point on the test but the rest I had no clue what to do!??!!

 

[jwpaine]

1/(x^3) * (y^8)^(1/3) =
1/(x^3) * (y^(8/3))/1 =
(y^(8/3))/x^3

 

[skeeter]

\(\displaystyle \L \sqrt[3]{y^8} = \sqrt[3]{y^3 \cdot y^3 \cdot y^2} = \sqrt[3]{y^3} \cdot \sqrt[3]{y^3} \cdot \sqrt[3]{y^2} = y \cdot y \cdot \sqrt[3]{y^2} = y^2 \cdot \sqrt[3]{y^2}\)

\(\displaystyle \L \sqrt[5]{y^7} = \sqrt[5]{y^5 \cdot y^2} = \sqrt[5]{y^5} \cdot \sqrt[5]{y^2} = y \cdot \sqrt[5]{y^2}\)

kapish?

 

[lovebugjay]

i think a little bit, this problem i missed all together it was (1/z)*√x^4z^15 i put the index was 2 i thought that would be the index...why was I wrong??

 

[Denis]

5√y^7 but the five needs to be above the √ like the three was i just don't know how to show that....sorry I am not very good at this stuff as you can see!!

Relax, Bug!
What that means is the 5th root of y^7;
next time, post like this: (y^7)^(1/5)

So answer is y^(7/5) : REMEMBER that (x^a)^b = x^(a * b)