#1

Okay i have a man cold and i'm confused by this so eah i've turned to UG

I'll just '/' to expressed square root signs

I have to express each of these in the form k /5

And i have /63 and 21 over /7

Any help? And then i have to hence write /63 + 21 over /7 in the form n /7 - which is 6 /7 but with the first bit i am lost

I'm mega confused

I'll just '/' to expressed square root signs

I have to express each of these in the form k /5

And i have /63 and 21 over /7

Any help? And then i have to hence write /63 + 21 over /7 in the form n /7 - which is 6 /7 but with the first bit i am lost

I'm mega confused

*Last edited by philipisabeast at Sep 30, 2007,*

#2

/63

= /9 * /7

=3/7

Easy enough

and 21 over /7

This one is harder, you have to rationalise the denominator. The only way to make /7 a rational number is to multiply it by /7.

So multiply 21 over /7 by /7 over /7

Then you will get 21/7 over 7 which will cancel down to 3/7 over 1, therefore 3/7 = answer.

= /9 * /7

=3/7

Easy enough

and 21 over /7

This one is harder, you have to rationalise the denominator. The only way to make /7 a rational number is to multiply it by /7.

So multiply 21 over /7 by /7 over /7

Then you will get 21/7 over 7 which will cancel down to 3/7 over 1, therefore 3/7 = answer.

#3

^^^ *stabs in the face*

I was so going to say that..

I was so going to say that..

#4

^^^ *stabs in the face*

I was so going to say that..

#5

AS level maths ftw!

#6

Laff it's AS but i had to give it in the form k /5 - he gave it in 7? Or is the question in fact sneaky?

#7

Laff it's AS but i had to give it in the form k /5 - he gave it in 7? Or is the question in fact sneaky?

Good point. Tell you what, I am doing some maths homework right now so I will have a blast at this mofo. Just wait for my edit.

EDIT 1: I have just found that 21 /7 over /7 (which we have already established to be 3 /7 goes to /9 /7 which goes to /63 so you essentially have /63 + /63 here.

EDIT 2: 63 into 5? wtf? This question phails to the root of epic.

*Last edited by MadClownDisease at Sep 30, 2007,*

#8

AS level maths ftw!

I'm having to do this at GCSE!

#9

I'm having to do this at GCSE!

The A-level ones aren't much harder...

#10

The A-level ones aren't much harder...

Yeah but I suck at maths. And there's no way I'll take it for A level.

#11

EDIT 2: 63 into 5? wtf? This question phails to the root of epic.

haha

I'll just leave it

#12

AS level maths ftw!

I don't recall doing surds at AS level.

#13

I just started my AS course and surds in in teh first unit...

#14

Yep it's alot of Core 1

#15

We havent touched on them at AS yet, but we have done them before in maths.

I'll have a go at these two though. Though the √63 one seems a bit odd

I'll have a go at these two though. Though the √63 one seems a bit odd

#16

I got a result but its a bit of a crap way to do it. (I did it in my head, so it might be wrong, but then again, C1 is non-calculator anyway)

√63 = √(3/5)*√105

= √(15/25)*√105

= √(15/25)*√21*√5

= √(315/25)*√5

= (√315)/5*√5

√63 = √(3/5)*√105

= √(15/25)*√105

= √(15/25)*√21*√5

= √(315/25)*√5

= (√315)/5*√5

#17

was wrong again.

this should be correct

√63+21 / √7

3√7+21 / √7

(3√7(√1/5)(√5))+(21(√1/5)(√5)) / √7

eliminate (√5) to find k

therefore k = (21√(1/5)+3√(7/5)) / √7

this should be correct

√63+21 / √7

3√7+21 / √7

(3√7(√1/5)(√5))+(21(√1/5)(√5)) / √7

eliminate (√5) to find k

therefore k = (21√(1/5)+3√(7/5)) / √7

*Last edited by xtremebigbird at Sep 30, 2007,*