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Mathematics Question

i want to simplify: 2(root 7) + 8 / (root 7) + 2
in the form p + q(root 7) where p and q are integers.

can i simplify the above expression to: (root7) + 4

is this correct?

Scroll to see replies

Reply 1

DanMushMan

BABYGUY

i want to simplify: 2(root 7) + 8 / (root 7) + 2
in the form p + q(root 7) where p and q are integers.

can i simplify the above expression to: (root7) + 4

is this correct?

Reply 2

theone

BABYGUY

i want to simplify: 2(root 7) + 8 / (root 7) + 2
in the form p + q(root 7) where p and q are integers.

can i simplify the above expression to: (root7) + 4

is this correct?

I'm gonna get this done quickly because I want to go to bed :biggrin:

2 root 7 + 8
--------------
root 7 + 2

= (2root7+8)(root7-2)
--------------------
(root7+2)(root7-2)

= (4root7 - 2)/3

Reply 3

BABYGUY

why????
how can i make 2(root 7) = 8 / (root 7) + 2
into form: p + q(root 7)

please help. :smile:

Reply 4

BABYGUY

yes but it says p + q(root 7)....your answer has 3 in a denominator, it doesnt want it to be a fraction.

Reply 5

theone

BABYGUY

yes but it says p + q(root 7)....your answer has 3 in a denominator, it doesnt want it to be a fraction.

Then divide each bit by 3 to get 4/3 root 7 - 2/3.

Reply 6

theone

elpaw

-6 + 4/5 rt 7

eh?

Reply 7

elpaw

theone

eh?

sorry wrote the numbers down wrong

Reply 8

Juwel

theone

Then divide each bit by 3 to get 4/3 root 7 - 2/3.

...What he said.

Aha

-2/3 + 4/3 rt 7 is what i get (when i write the numbers down right! :biggrin:

)

Reply 11

DanMushMan

BABYGUY

i want to simplify: 2(root 7) + 8 / (root 7) + 2
in the form p + q(root 7) where p and q are integers.

can i simplify the above expression to: (root7) + 4

is this correct?

thought you were 'too good for oxbridge' well, good luck

Reply 12

Juwel

DanMushMan

thought you were 'too good for oxbridge' well, good luck

Hahaha! Sure told him!

Reply 13

theone

elpaw

-2/3 + 4/3 rt 7 is what i get (when i write the numbers down right! :biggrin:

)

Can you go through your working?

Reply 14

Unregistered

p and q are supposed to be integers though???

Reply 15

DanMushMan

theone

Can you go through your working?

you r correct

Reply 16

theone

Unregistered

p and q are supposed to be integers though???

No - it's not possible. Stupid question by a stupid poster.

Reply 17

elpaw

theone

Can you go through your working?

(8 + 2 rt 7)/(2 + rt 7) = ((8 + 2 rt 7)(2 - rt 7))/((2 + rt 7)(2 - rt 7))
= (16 - 8 rt 7 + 4 rt 7 - 14)/(4 - 2 rt 7 + 2 rt 7 - 7)
= (2 - 4 rt 7)/(-3)
=-2/3 + 4/3 rt 7

Reply 18

Juwel

elpaw

(8 + 2 rt 7)/(2 + rt 7) = ((8 + 2 rt 7)(2 - rt 7))/((2 + rt 7)(2 - rt 7))
= (16 - 8 rt 7 + 4 rt 7 - 14)/(4 - 2 rt 7 + 2 rt 7 - 7)
= (2 - 4 rt 7)/(-3)
=-2/3 + 4/3 rt 7

<Puffs cheeks> Good stuff.

Reply 19

BABYGUY

lol.........i love winding you up. i give these pointless maths threads where i already have the answer. and the funnire thing is, i lied about the integers thing, i said rational numbers lol..omg.you lot fell for not rationalising the denominator.hilarious. omg. funny. but i love hearing you lot squabble over these maths problems!!!! rember the last thread!! 10 pages long and i already knew the answer!!!