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ah, now ive just stumbled upon another surd question.
(√a+√b)²=a+b+2√ab
determine the square root of 17+6√8
can someone explain how to do this please?
I hope you can help,
thanks.
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(±(√a+√b))²=a+b+2√ab
the right side your squared term, the left side is your root squared.
so what you have to do is put 17+6√8 into the form a+b+2√ab
17+6√8 = 17+(2×3)√8 = 17+2√(8×9)
note now that, 17 = 8+9. and this is in the required form.
(√8+√9)²=8+9+2√(8×9)
which gives. √(17+6√8) = ±(√8 + √9)
Last edited by luca-deltodesco (2008-04-04 18:58:10)
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Thanks luca:D
"If your going through hell, keep going."
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Here is a fun one. What is:
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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luca, the answer is NOT ±(√8 + 3), square roots are always positive
Last edited by Kurre (2008-04-06 04:23:22)
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But what's all this thing about my Maths Teacher saying next year we'd be doing stuff on imaginary numbers? Like square roots of negative numbers?
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
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It's not too hard to understand imaginary numbers if you're good with surds.
Basically a new number "i" is introduced and defined to be √(-1), and all the same surd rules apply after that.
So for example, √(-9) =√9 x √(-1) = 3i.
Perhaps more importantly, i² = -1. That means imaginary things can sometimes combine and become real again.
For example, (4+i)(4-i). [Note - since those are real and imaginary numbers added together, I should technically be calling them 'complex']
Expand that like you would normally and you get 4² + 4i - 4i - i² = 4² - i² = 17.
Then later on you start representing imaginary numbers as a product of exponents and trig functions and other crazy things like that. It's great fun.
Why did the vector cross the road?
It wanted to be normal.
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Oooh! Sounds it.
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
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