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Did i rationalize this correctly?

http://prntscr.com/awxuxy

Is this correct?

Or was the demoninator ment to be 3? because root 2 multiply by root 2 gives root 4 and the root of 4 is = 2 so then do you do 2 + 1 ? or 2 x 1?

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Reply 1
Avatar for nerak99
nerak99
13
No. so far as I can tell you have multiplied up and down by root 2. In this case you multiply up and down by the conjugate of the bottom. (i.e. swap the sign on the root bit to give 1 - root 2).

This works because it all depends on the difference between two squares which causes the root bit to disappear from the denominator when multiplied out hence 51+2=5(1−2)(1+2)(1−2) \frac{5}{1+\sqrt 2}=\frac{5(1-\sqrt 2)}{(1+\sqrt 2)(1-\sqrt 2)}

=5(1−2)1−2=5(1−2)−1=−5(1+2) =\frac{5(1-\sqrt 2)}{1-2}=\frac{5(1-\sqrt 2)}{-1}=-5(1+\sqrt 2)

Which I would leave in that form although if the Q asked for it to be fully expanded would become

−5−52) -5-5\sqrt 2)
Reply 2
Avatar for junayd1998
junayd1998
OP
11
Original post by nerak99
No. so far as I can tell you have multiplied up and down by root 2. In this case you multiply up and down by the conjugate of the bottom. (i.e. swap the sign on the root bit to give 1 - root 2).

This works because it all depends on the difference between two squares which causes the root bit to disappear from the denominator when multiplied out hence 51+2=5(1−2)(1+2)(1−2) \frac{5}{1+\sqrt 2}=\frac{5(1-\sqrt 2)}{(1+\sqrt 2)(1-\sqrt 2)}

=5(1−2)1−2=5(1−2)−1=−5(1+2) =\frac{5(1-\sqrt 2)}{1-2}=\frac{5(1-\sqrt 2)}{-1}=-5(1+\sqrt 2)

Which I would leave in that form although if the Q asked for it to be fully expanded would become

−5−52) -5-5\sqrt 2)


I thought when your rationalize you have to multiply by the denominator.. Im a bit confused now, because I just watched a video where is shows you how to rationalize the denominator and it says you multiply the root at the bottom (denominator) at both top and bottom. ugh
Reply 3
Avatar for nerak99
nerak99
13
That only applies where the denominator is only a root and not of the form a+b a+ \sqrt b.


Hence: rationalise ab  ⇒abbb=abb \frac {a}{\sqrt b}\;\Rightarrow \frac{a \sqrt b}{\sqrt b \sqrt b}=\frac {a \sqrt b}{b}
(edited 7 years ago)
Reply 4
Avatar for junayd1998
junayd1998
OP
11
Original post by nerak99
That only applies where the denominator is only a root and not of the form a+b a+ \sqrt b.


Hence: rationalise ab  ⇒abbb=abb \frac {a}{\sqrt b}\;\Rightarrow \frac{a \sqrt b}{\sqrt b \sqrt b}=\frac {a \sqrt b}{b}


So what do you do when it is that form? Sorry im just confused
Reply 5
Avatar for Zacken
Zacken
22
Original post by junayd1998
So what do you do when it is that form? Sorry im just confused


You have two forms:

1. ab\displaystyle \frac{a}{\sqrt{b}} - in this case, you multiply top and bottom by b\sqrt{b}.

2. ab+c\displaystyle \frac{a}{b + \sqrt{c}} - in this case, you multiply top and bottom by b−cb - \sqrt{c} (i.e: you flip the sign next to the square root, if it's a + it becomes -, if it's a - it becomes a +).
Reply 6
Avatar for nerak99
nerak99
13
Original post by junayd1998
So what do you do when it is that form? Sorry im just confused

By "that" I assume you mean a a in the numerator and b \sqrt b in the denominator and in that case you can just multiply top and bottom by b \sqrt b .

Example 23=2333=233\frac {2}{\sqrt 3}=\frac{2 \sqrt 3}{\sqrt 3 \sqrt 3}=\frac {2\sqrt 3}{3}

If the form is x5−2=x(5+2)(5−2)(5+2)=x(5+2)5−2=x(5+23 \frac{x}{5-\sqrt 2}=\frac {x(5+\sqrt 2)}{(5-\sqrt 2)(5+\sqrt 2)}=\frac{x(5+\sqrt 2)}{5-2}=\frac{x(5+\sqrt 2}{3}

the x in the numerator is any bit of algebra you like
Reply 7
Avatar for junayd1998
junayd1998
OP
11
Original post by Zacken
You have two forms:

1. ab\displaystyle \frac{a}{\sqrt{b}} - in this case, you multiply top and bottom by b\sqrt{b}.

2. ab+c\displaystyle \frac{a}{b + \sqrt{c}} - in this case, you multiply top and bottom by b−cb - \sqrt{c} (i.e: you flip the sign next to the square root, if it's a + it becomes -, if it's a - it becomes a +).


Ah i understand never knew the second form, I was only aware of the first one. Thanks i get it now
Reply 8
Avatar for junayd1998
junayd1998
OP
11
Original post by Zacken
You have two forms:

1. ab\displaystyle \frac{a}{\sqrt{b}} - in this case, you multiply top and bottom by b\sqrt{b}.

2. ab+c\displaystyle \frac{a}{b + \sqrt{c}} - in this case, you multiply top and bottom by b−cb - \sqrt{c} (i.e: you flip the sign next to the square root, if it's a + it becomes -, if it's a - it becomes a +).


http://prntscr.com/ax56l2

Have I done this correctly? I'm a bit unsure.
Reply 9
Avatar for Zacken
Zacken
22
Original post by junayd1998
http://prntscr.com/ax56l2

Have I done this correctly? I'm a bit unsure.


Perfect, now simplify the 1-2.
Original post by Zacken
Perfect, now simplify the 1-2.




http://prntscr.com/ax5bng

That any good?
Reply 11
Avatar for Zacken
Zacken
22
Original post by junayd1998


Nooo, 5 - sqrt(2) is not 5sqrt(2). Instead (5 - sqrt(2)) / -1 = sqrt(2) - 5
Original post by Zacken
Nooo, 5 - sqrt(2) is not 5sqrt(2). Instead (5 - sqrt(2)) / -1 = sqrt(2) - 5


http://prntscr.com/ax607w

So that? The other way round?
Reply 13
Avatar for Zacken
Zacken
22
Original post by junayd1998
http://prntscr.com/ax607w

So that? The other way round?


Perfect. :smile:
Original post by Zacken
Perfect. :smile:


Can i ask why we do it that way round? Because with a question like this http://prntscr.com/ax63xb the number comes first then the root with the number.:colondollar: Sorry for the annoyance just want to make sure I know it all correctly
Reply 15
Avatar for Zacken
Zacken
22
Original post by junayd1998
Can i ask why we do it that way round? Because with a question like this http://prntscr.com/ax63xb the number comes first then the root with the number.:colondollar: Sorry for the annoyance just want to make sure I know it all correctly


There is a massive difference between aba\sqrt{b} which is the number aa multiplied by the number b\sqrt{b} and a+ba + \sqrt{b} which is the number aa added to the number b\sqrt{b}.

If you want, you could have written −5+2=2−5-5 + \sqrt{2} = \sqrt{2} - 5 (they mean the same thing) but they are very different from 525\sqrt{2} or −52-5\sqrt{2}.

It's like you saying that 1+11 + 1 is the same thing as 1×11 \times 1.
Original post by Zacken
There is a massive difference between aba\sqrt{b} which is the number aa multiplied by the number b\sqrt{b} and a+ba + \sqrt{b} which is the number aa added to the number b\sqrt{b}.

If you want, you could have written −5+2=2−5-5 + \sqrt{2} = \sqrt{2} - 5 (they mean the same thing) but they are very different from 525\sqrt{2} or −52-5\sqrt{2}.

It's like you saying that 1+11 + 1 is the same thing as 1×11 \times 1.


Oh okay I understand Thanks
Reply 17
Avatar for Zacken
Zacken
22
Original post by junayd1998
Oh okay I understand Thanks


Awesome.
The correct answer is, -5 +5√2
Reply 19
Avatar for Zacken
Zacken
22
Original post by junayd1998
Oh okay I understand Thanks


As above - (I have no clue how this slipped my eye) but the correct answer should be 5(1−2)−1=5−52−1=52−5\frac{5(1 - \sqrt{2})}{-1} = \frac{5 - 5\sqrt{2}}{-1} = 5\sqrt{2} - 5, i.e: you need to multiply out the 5(1−2)5(1-\sqrt{2}) as 5−525 - 5\sqrt{2} and not as 5−25 - \sqrt{2}.

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