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simplifying in the form a+b?c ??

I've just been given a question:

Simplify the following in the form a + b?c where a, b and c are intergers:

a) (3+?2)^2

b) (2?3+4)(2-?3)

c) 2/?3-1

No need to actually give me the answers, just tell me how to simplify them :smile:

Must have been sleeping/generally not paying attention when I was supposed to learn these :rolleyes: Or it could be that these questions were thrown in just to be evil and I was never taught them at all! :eek:

(probably the former :colondollar:)

Thanks in advance.
(edited 13 years ago)
Reply 1
OK, I don't know why my square root signs are showing up as question marks, sorry.
Original post by james-brown28
I've just been given a question:

Simplify the following in the form a + b?c where a, b and c are intergers:

a) (3+?2)^2

b) (2?3+4)(2-?3)

c) 2/?3-1

No need to actually give me the answers, just tell me how to simplify them :smile:

Must have been sleeping/generally not paying attention when I was supposed to learn these :rolleyes: Or it could be that these questions were thrown in just to be evil and I was never taught them at all! :eek:

(probably the former :colondollar:)

Thanks in advance.


For a and b, just expand the brackets and collect like terms. c is already in that form ( or it could be written -1+2rt3)
Reply 3
Original post by james-brown28
I've just been given a question:

Simplify the following in the form a + b?c where a, b and c are intergers:

a) (3+?2)^2

b) (2?3+4)(2-?3)

c) 2/?3-1

No need to actually give me the answers, just tell me how to simplify them :smile:

Must have been sleeping/generally not paying attention when I was supposed to learn these :rolleyes: Or it could be that these questions were thrown in just to be evil and I was never taught them at all! :eek:

(probably the former :colondollar:)

Thanks in advance.


Use this: (a+b) x (c+d) = ac+ad+bc+bd

Also use 1/sqr(3) = sqr(3)/3

where sqr is square root.

So (3+sqr(2))^2 = 9 + 2 + 6sqr(2) = 11 + 6sqr(2)
Reply 4
a) Use the fact that

(p+q)2=(p+q)(p+q)(p+\sqrt{q})^2 = (p+\sqrt{q})(p+\sqrt{q})
(p+q)(p+q)=p2+2pq+q(p+\sqrt{q})(p+\sqrt{q}) = p^2 + 2p\sqrt{q} + q

b) Use similar ideas to above.

c) Rationalise the denominator.

a+bcd×c+dc+d=(a+b)(c+d)(cd)(c+d)\frac{a+\sqrt{b}}{c-\sqrt{d}} \times \frac{c+\sqrt{d}}{c+\sqrt{d}} = \frac{(a+\sqrt{b})(c+\sqrt{d})}{(c-\sqrt{d})(c+\sqrt{d})}
Reply 5
Original post by indifferencepersonified
For a and b, just expand the brackets and collect like terms. c is already in that form ( or it could be written -1+2rt3)


No, the sqr ia at the bottom in c

2/?3-1 is in fact -1 + 2/3 sqr(3)
Reply 6
Original post by Ienjoylife
No, the sqr ia at the bottom in c

2/?3-1 is in fact -1 + 2/3 sqr(3)


No.

2311+233\frac{2}{\sqrt{3}-1} \neq -1 + \frac{2}{3} \sqrt{3}
Original post by Ienjoylife
No, the sqr ia at the bottom in c

2/?3-1 is in fact -1 + 2/3 sqr(3)


I misread that, I didn't see the '/'
Reply 8
Original post by Noble.
No.

2311+233\frac{2}{\sqrt{3}-1} \neq -1 + \frac{2}{3} \sqrt{3}


errmm excuse me but there were no brackets in the original stem (c). It showed as 231\frac{2}{\sqrt{3}}-1
Reply 9
Original post by Ienjoylife
errmm excuse me but there were no brackets in the original stem (c). It showed as 231\frac{2}{\sqrt{3}}-1


I'm fairly sure the question was a bog-standard rationalise the denominator question. Kind of pointless for 231\frac{2}{\sqrt{3}}-1

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