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# Mathematics Question watch

1. 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?
2. (Original post by 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?
no
3. (Original post by 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

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

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

= (4root7 - 2)/3
4. why????
how can i make 2(root 7) = 8 / (root 7) + 2
into form: p + q(root 7)

5. yes but it says p + q(root 7)....your answer has 3 in a denominator, it doesnt want it to be a fraction.
6. (Original post by 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.
7. (Original post by elpaw)
-6 + 4/5 rt 7
eh?
8. (Original post by theone)
eh?
sorry wrote the numbers down wrong
9. (Original post by theone)
Then divide each bit by 3 to get 4/3 root 7 - 2/3.
...What he said.
10. Aha
11. -2/3 + 4/3 rt 7 is what i get (when i write the numbers down right! )
12. (Original post by 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
13. (Original post by DanMushMan)
thought you were 'too good for oxbridge' well, good luck
Hahaha! Sure told him!
14. (Original post by elpaw)
-2/3 + 4/3 rt 7 is what i get (when i write the numbers down right! )
Can you go through your working?
15. p and q are supposed to be integers though???
16. (Original post by theone)
Can you go through your working?
you r correct
17. (Original post by Unregistered)
p and q are supposed to be integers though???
No - it's not possible. Stupid question by a stupid poster.
18. (Original post by 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
19. (Original post by 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.
20. 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!!!

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Updated: November 5, 2003
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