Question

Please help me :)
meoww?

Answer 1

let \[\sqrt{\frac{ 7 }{ 12 }- \frac{ \sqrt{3} }{ 3 }} = x - \sqrt{y}\]
square both sides and form a pair of simultaneous equations to find x and y given they are rational. hence find (simplified form)
\[\sqrt{\frac{ 7 }{ 12 }- \frac{ \sqrt{3} }{ 3 }}\]

Answer 2

@Kuoministers

Start by squaring both sides and show me your working,I can help you where you go wrong.

Answer 3

sure

Answer 4

Use


(x-y)^2=x^2-2ab+b^2

When you solve the RHS of your equation

Answer 5

I mean

(x-y)^2=x^2-2xy+y^2*

Answer 6

\[\frac{ 7 }{ 12 } - \frac{ \sqrt{3} }{ 3 } = x^{2} - 2x \sqrt{7} + y\]

Answer 7

Where did you get sqrt{7} from ?

Answer 8

\[\frac{ 7 }{ 12 } = x^{2} + y\]
\[\frac{- \sqrt{3} }{ 3 } = -2x \sqrt{y}\]
ohh that supposed to be y

Answer 9

i dont know what to do :p

Answer 10

For the second equation, express x in terms of y.

Answer 11

\[(-\frac{ \sqrt{3} }{ 3 } + 2x)^{2} = y\]
right?

Answer 12

nono i cant do that

Answer 13

im so lost...

Answer 14

Just consider the second equation.
Divide both sides by \(-2\sqrt y\) . What do you get?

Answer 15

yea i dont know how to divide the RHS by that \[\sqrt[-2]{y}\]

Answer 16

\[\frac{- \sqrt{3} }{ 3 } = -2x \sqrt{y}\]\[\frac{\frac{- \sqrt{3} }{ 3 }}{-2 \sqrt{y}} = \frac{-2x \sqrt{y}}{-2\sqrt{y}}\]Cancel what you can cancel.Simplify it, what do you get?

Answer 17

umm the negative on the RHS i think?
idk what on the right hand side... but the left hand side its x
:P

Answer 18

left right :|

You should get +ve for both sides :|

Answer 19

:P wow im a special child :O

Answer 20

anyways i dont know how to simplify that

Answer 21

\[\frac{\frac{- \sqrt{3} }{ 3 }}{-2 \sqrt{y}} = \frac{-2x \sqrt{y}}{-2\sqrt{y}}\]\[\frac{- \sqrt{3} }{ 3(-2) \sqrt{y} } = x\]\[\frac{\sqrt{3} }{ 6\sqrt{y} } = x\]Got it?

Answer 22

oh yea yea i got that!!

Answer 23

Sub. this into the first equation, what do you get?

Answer 24

\[\frac{ 7 }{ 12 } = (\frac{ \sqrt{3} }{ 6\sqrt{y} })^{2} + y\]

Answer 25

Okay, simplify the *right* side

Note: left = right

Answer 26

\[\frac{ 7 }{ 12 } = \frac{ 3 }{ 36y } + y\]

Answer 27

Simplify the first term on the right.

Answer 28

\[21y = 3 + 36y ^{2}\]

Answer 29

Can you solve the quadratic?

Answer 30

ofc

Answer 31

(9y - 3)(4y - 1) right?

Answer 32

(9y - 3)(4y - 1)=0

Answer 33

so whats the final answer?

Answer 34

answer in book says \[-\frac{ 1 }{ 2 } + \frac{ \sqrt{3} }{ 3 }\]

Answer 35

i think our answer for y was wrong?

Answer 36

What's your answer for y?

Answer 37

root 3 \ 3

Answer 38

i think we didnt rationalise the denominator?

Answer 39

:\
(9y - 3)(4y - 1)=0
(9y - 3)=0 or (4y - 1)=0
y = 1/3 or y = 1/4

Put y=1/3 into \(\frac{\sqrt{3} }{ 6\sqrt{y} } = x\)
x = ...?

Put y = 1/4 into \(\frac{\sqrt{3} }{ 6\sqrt{y} } = x\)
x = ...?

Answer 40

x= half
x = sqrt 3 over half

Answer 41

Second one is not quite right :|

Answer 42

sqrt 3 over 3?

Answer 43

Yup, so now, you have two pairs of solutions,
\((\frac{1}{2} , \frac{1}{3})\) and \((\frac{\sqrt{3}}{3}, \frac{1}{4})\)

Consider\(\sqrt{\frac{ 7 }{ 12 }- \frac{ \sqrt{3} }{ 3 }} = x - \sqrt{y}\)
Pick the suitable pair now!

Answer 44

i cant read it, too small... 1 over what?

Answer 45

\[(\frac{1}{2}, \frac{1}{3})\] and \[(\frac{\sqrt{3}}{3}, \frac{1}{4})\]

Answer 46

aint both pair suitable? ill pick 1 over 3 and 1 over 4?

Answer 47

\[\huge\ \sqrt{\frac{ 7 }{ 12 }- \frac{ \sqrt{3} }{ 3 }} = x - \sqrt{y}\]

Answer 48

No, they are not the same. When you take the square root, the right side should be +ve.
One pair gives you a -ve result. Check it~

Answer 49

oh yea... so answer would be the half and sqrt 3 over 3

Answer 50

Wait *~*

Answer 51

\[\frac{ 1 }{ 2 } - \frac{ \sqrt{3} }{ 3 }\]

Answer 52

no?

Answer 53

Check to see if it is +ve or -ve :|

Answer 54

thats negative

Answer 55

Can we have a negative number after taking the square root? :|

Answer 56

ofc not :P

Answer 57

So, that is wrong :|

Answer 58

so what would it be?

Answer 59

What is the other pair?

Answer 60

1r4 1r3

Answer 61

\[(\frac{1}{2}, \frac{1}{3})\]

\[(\frac{\sqrt{3}}{3}, \frac{1}{4})\]
Which one was the pair you'd just tried and it was wrong?

Answer 62

Top one or bottom one?

Answer 63

umm top?

Answer 64

Yup, so the bottom one is the correct one :|

Answer 65

ahh ok..

Answer 66

Do... you ... understand???

Answer 67

yea

Answer 68

Okay :)

Answer 69

\[{\color{blue}{\rightarrow \boxed{\mathbb{\text{THANK YOU SO MUCH FOR YOUR TIME}}}}} \]

Answer 70

like a hours worth :P

Answer 71

You're welcome! (Nice LaTeX! )
Sorry that I didn't replied promptly :( I'm not feeling well and my brain was(is) working very slow!