Surds

OP

distortedgav

Square both sides, then try to prove that (LHS)^2 = (RHS)^2

So you start with the (LHS)^2 and basically expand brackets until you get down to the answer 2, which is the (LHS)^2.

That's basically how you do it.

Thats what I thought straight way, but I am having difficulties doing LHS², not really sure how to treat it.
Can you expand it for me?

Reply 3

-G-a-v-

16

√(2+√3) - √(2-√3) = √2

Squaring:

[√(2+√3) - √(2-√3)]² = 2, and that's really what you need to prove since if that is true, the first line must be true.

Start with the LHS^2

LHS^2=(√(2+√3) - √(2-√3))(√(2+√3) - √(2-√3))
=√(2+√3) [√(2+√3) - √(2-√3)] - √(2-√3)[√(2+√3) - √(2-√3)]
= √(2+√3)√(2+√3) - √(2+√3)√(2-√3) - √(2-√3)√(2+√3) + √(2-√3)√(2-√3)
= 2 + √3 - √[(2+√3)(2-√3)] - √[(2-√3)(2+√3)] + 2 - √3
= 2 + √3 - √(4-3) - √(4-3) + 2 - √3
= 2 + √3 - 1 - 1 + 2 - √3
= 4 - 2
= 2 = RHS^2

So by square rooting both sides, you get back to LHS = RHS, and you're done :smile:

....I think :p:

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It's just basically knowing basic surd laws i.e. root (ab) = (root a)(root b) and knowing the difference of two squares, and expanding a lot of brackets. It's not the nicest of questions, never seen it asked before though, where's the question from?

Reply 4

LHS² = 2+√3 - 2√(2+√3)√(2-√3) + 2 - √3

= 4 - 2√[(2+√3)(2-√3)]

and expand out under the square root to give LHS² = 4-2 = 2 = RHS²

Reply 5

-G-a-v-

16

Malik

Thats what I thought straight way, but I am having difficulties doing LHS², not really sure how to treat it.
Can you expand it for me?

I know I'm multiposting quite a bit now lol, but the best way to treat anything like this is just to work through it slowly, that way you make far fewer mistakes.

Reply 6

OP

Thanks gav owe you one, must taken a while to type out. I was afraid it might be quite messy, and It seems i was right.

Its from a Maths AEA paper and its only part(a) :redface:

.

Reply 7

-G-a-v-

16

..unless of course you're Kaiser Mole and are just plain quick to see things like this :p:

lol

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Malik

Thanks gav owe you one, must taken a while to type out. I was afraid it might be quite messy, and It seems i was right.

Its from a Maths AEA paper and its only part(a) :redface:

.

In which case they'd be probably looking to see if you could spot a quicker way, like I didn't :p:

... wasn't really difficult to type out, just did a lot of copying and pasting since it all looks the same anyway :smile:

Reply 8

I haven't posted on maths forums much for a while now, very odd, just eager to write some equations down again :p:

Reply 9

OP

KAISER_MOLE

I haven't posted on maths forums much for a while now, very odd, just eager to write some equations down again :p:

lol spoken like a true Mathematician!

Reply 10

Esquire

1

ok ... so lets look at the LHS, I will make it =2.
LHS^2=[√(2+√3)-√(2-√3)]^2 ...consider (a+b)^2=a^2+2ab+b^2
= [2+√3]-[2√(2-√3)√(2+√3)]+[2-√3] ...now...√a√b=√ab...middle bit
=4-[2√(2-√3)(2+√3)]
=4-[2√(4-3)]
=4-2
=2 as required

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erff sorry i'm a bit drunk... oh well too slow.

Reply 11

Malik

lol spoken like a true Mathematician!

trying to get back on the good stuff!

Reply 12

topcrabcakes

11

By proving (LHS)²=(RHS)² doesn't prove the identity, otherwise -2=2 e.g.?

Reply 13

OP

wacabac

By proving (LHS)²=(RHS)² doesn't prove the identity, otherwise -2=2 e.g.?

I was thinking of it more the lines of

(LHS)²=(RHS)²
±2=±2

Reply 14

Neapolitan

2

wacabac

By proving (LHS)²=(RHS)² doesn't prove the identity, otherwise -2=2 e.g.?

Yes it does as the LHS is obviously positive

Reply 15