Question

MEDAL AND FANN
i need help with simplifying rational expressions
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Answer 1

@SolomonZelman

Answer 2

@phi

Answer 3

Have you ever seen this "rule"?
\[ \sqrt[6]{x} = x^\frac{1}{6}\]

Answer 4

well yes but i dont know how to use it ..

Answer 5

i have to rationalize the denominator in the expression..

Answer 6

write your problem as
\[ \sqrt[6]{8a^6} = \left(8a^6 \right)^\frac{1}{6} \]
and it is good if you notice that 8 is 2*2*2 (when doing these problems, you should expect to factor numbers to find this kind of relation)

2*2*2= 2^3

so you have
\[ \left(2^3a^6 \right)^\frac{1}{6} \]

Answer 7

now use this useful rule:
\[ ( a\cdot b)^c = a^c b^c \]

Answer 8

\[2^{1/2} * a\]

Answer 9

and finally (if the answer choices use the radical version)
rewrite 2^(½) as \( \sqrt{2} \)
to get
\[ a \sqrt{2} \]

Answer 10

notice for square roots we do not bother to write
\[ \sqrt[2]{2} \]
but just
\[ \sqrt{2} \]

Answer 11

it says it was incorrect..

Answer 12

what did you enter ?

Answer 13

you left off the "a"

Answer 14

ohh.. i have to move on to the next problem.. can you help me with that ?

Answer 15

and finally (if the answer choices use the radical version)
rewrite 2^(½) as \( \sqrt{2}\)
to get
\[ a \sqrt{2} \]

Answer 16

same idea

write 125 as 5*5*5 = 5^3
and write the problem as
\[ \left(5^3 a^{15}\right)^\frac{1}{6} \]

Answer 17

it might be easier if you write a^15 as a^12 * a^3

Answer 18

\[5^{1/2} * a ^{5/2}\]

Answer 19

ok, but write a^(5/2) as a^(4/2) * a^(½)

Answer 20

so \[5a ^{1/2} * a ^{4/2} * a ^{1/2}\]

Answer 21

or is that wrong ?

Answer 22

5^½∗a^5/2 = 5^(½) a^(4/2) a^(½)

notice you left off the ½ exponent on the 5

also, we can simplify 4/2 to just 2

to get
5^(½) a^2 a^(½)

Answer 23

using radicals:
\[ \sqrt{5} a^2 \sqrt{a} \\
a^2 \sqrt{5}\sqrt{a} \]
but normally we would "combine" the square roots
\[ a^2 \sqrt{5a} \]
it looks nicer

Answer 24

so is that the simplified version?

Answer 25

@phi thanks ! can you help me with 2 more please ?

Answer 26

Can you make them a new post please?