- anonymous

rationalize denominator : please help
1.) 4/ sqrt. 7p
2.) srqt.98/x
3.) 9/3 sqrt.y
4.) sqrt.7 / sqrt.5 +7

- jamiebookeater

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- anonymous

1.) \[4/ \sqrt{7p}\]
2.) \[\sqrt{98}/x\]
3.) \[9/ 3\sqrt{y}\]
4.) \[\sqrt{7}/ \sqrt{5} + 7\]

- anonymous

please hlep , i dont understand them

- dumbcow

ok the idea is to get the sqrt out of the denominator
For 1) multiply by sqrt(7p)/sqrt(7p)

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## More answers

- dumbcow

sqrt(x) * sqrt(x) = x

- anonymous

so it would be p ?

- dumbcow

7p everything within the radical
sorry imagine if i said x = 7p
->4*sqrt(7p)/7p
that is the rationalised form for the first one

- dumbcow

For number 4 you have to use a conjugate which is basically the denominator except you flip the sign
->sqrt(5)-7
multiply this on top and bottom by distributing (using FOIL)
and the sqrts in denominator will go away

- dumbcow

number 2 is already rationalised but you can simplify sqrt(98) by factoring out a perfect square
number 3 can be done same way as num 1, dont forget to reduce
hope this helps

- anonymous

so how would it be done ?

- anonymous

and thank you for helping , im just really bad in math

- dumbcow

which part are you stuck on

- dumbcow

figure it out??

- anonymous

all of it , especially foil

- dumbcow

ok lets look at num 2
since there is no sqrt in denominator, we don't have to do anything there
but we can simplify sqrt(98)
Factors of 98 = 2*49 =2*7*7
rewrite sqrt(98) as sqrt(2)*sqrt(7)*sqrt(7)
remember sqrt(7)*sqrt(7) = 7
leaving 7sqrt(2)

- anonymous

7sqrt. 2x

- anonymous

?

- anonymous

for number one i got : 4sqrt.7p/7p ? correct ?

- dumbcow

7sqrt(2)/x
sorry no the x doesnt change it stays on the bottom

- dumbcow

correct

- dumbcow

for num 3 we want to get rid of sqrt(y) by multiplying by sqrt(y) on top and bottom
->9/3sqrt(y) * sqrt(y)/sqrt(y) = 9sqrt(y)/3y
Now reduce
what does 9/3 reduce to?

- anonymous

or would 7p cancel itself ?

- anonymous

to 3

- dumbcow

correct
ohh no the 7p wouldnt cancel because the one on top is inside the sqrt, that is important

- dumbcow

answer for 3) is 3sqrt(y)/y

- anonymous

oh no your answer for 3 is wrong , it doesnt come out in my exam as an option

- dumbcow

hmm maybe the sqrt is in the numerator
try 3sqrt(y)

- anonymous

there is no 3 , only 9 sqrt

- dumbcow

hmm maybe the sqrt is in the numerator
try 3sqrt(y)

- dumbcow

interesting, what are the possible answers they give

- anonymous

number 3 : A) 9^3 sqrt.y^2 /y
b.) 9y
c.) 9^3 sqrt. y^2
d.) 9^3 sqrt. y /y

- dumbcow

and your sure the question was 9/(3sqrt(y))

- anonymous

yes

- anonymous

it better view higher above ^^^^ with the equation button

- anonymous

all the way up^^^^^^^^^^^^^^^^^^^^^^

- dumbcow

well in that case the answer is 3sqrt(y)/y
i'd bet on d) , weird you might have to talk to teacher on this one

- anonymous

and for the rest of the problems ?

- dumbcow

well what about the first 2 are those answers available :)

- anonymous

do you know how to solve this though ?

- dumbcow

rationalise fractions with sq roots? yes i know how to solve those but maybe they're looking for something else
look at prev examples and see if it lines up with what i've shown

- anonymous

correction : \[9 / ^{3} \sqrt{y}\]

- anonymous

so it changes everything ?

- dumbcow

little bit
the answer is A
think of it now as fractional exponents
cube root of y = y^(1/3)
to cancel that we multiply top and bottom by y^2/3
->y^1/3 * y^2/3 = y (add the exponents)
leaving (9*y^2/3)/y

- anonymous

you mean : 9^3 sqrt.y^2 / y ?

- dumbcow

oh y^2/3 can be written as cube root of y^2

- anonymous

its the same than ?

- dumbcow

yes

- anonymous

thank youuu ! its so complicated to me. and the rest ?

- dumbcow

oh yes FOIL is just way to remember how to distribute.
what is result of (x+2)*(x-3)?
_>distribute the x to both the x and -3
->x*x + x*-3 = x^2 - 3x
->then distribute the 2 to the x and -3
->2*x + 2*-3 = 2x -6
put it together
->x^2 -3x +2x -6 = x^2 -x -6
make sense?

- anonymous

is there an easier way ?

- anonymous

i have to use (x+2)*(x-3) in the equation to cancel out sqrt ?

- dumbcow

no that was just an example
and not really you still have to do the multiplications
For this problem we would be multiplying
(sqrt(5)+7) * (sqrt(5) - 7)

- dumbcow

sorry i gotta go
the answer is (7sqrt(7) - sqrt(35))/44
try to figure it out on your own by multiplying what i put above for denominator
then multiply sqrt(7)*(sqrt(5)-7) for numerator
Then at the end you have to distribute a negative by flipping the signs
good luck

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