Radical equation

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19 comments, last by alvaro 10 years, 5 months ago

Whenever you square both sides of an equation, you might be introducing false solutions, where the two sides of the equal have opposite signs. That's why at the end of this type of manipulation you need to verify that the values you found are actually solutions.

Squaring both sides of an equation is valid; squaring the terms of both sides is not, however.

But that's not what he did... He correctly deduced `1 = 49 * x' from `-1 = 7 * sqrt(x)'.

No, that's not a correct deduction. That's squaring each term, not the squaring entire side. When you square each term like that, you are assuming that ?(zw) = ?z?w, which is not generally true.

He said he squared each side, but then squared each term, not each side. That's the issue he had.

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sqrt(zw) does equal sqrt(z)*sqrt(w) as long as both z and w are positive or zero. Negative numbers => complex numbers mess things up.

(a+b)2 != a2 + b2 though (binomial theorem says otherwise. Pascal's triangle and such), unless in very special circumstances which don't occur in gamedev much or at all ("idiot binomial theorem" something to do with number theory and modulo arithmetic, google fu is failing only links I could find are in massive pdfs about Galois theory which is abstract algebra and gets the Paradigm Shifter seal of approval for hardcore math(s)).

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

When in doubt, ask wolfram biggrin.png

http://www.wolframalpha.com/input/?i=3%E2%88%9Ax+%2B+8+%3D+10%E2%88%9Ax+%2B+9+

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Looks like it, sqrt(z) + 1/7 is holomorphic so takes all except 2 values in the complex numbers including complex infinity, i.e. the Riemann sphere, I think ;)

It just happens 0 isn't one of the values it never has...

http://www.wolframalpha.com/input/?i=sqrt%28z%29+%2B+1%2F7+

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

No, that's not a correct deduction. That's squaring each term, not the squaring entire side.


You keep saying that, and yet, the square of -1 is 1 and the square of 7*sqrt(x) is 49*x. I don't want to appeal to authority, but I am a professional mathematician, and I am having a very hard time understanding what's wrong with that deduction.

I think the confusion arrives from a2 = b2 does not mean a = b

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

Or that (7*sqrt(x))2 just so happens to equal 49*x... but (7+sqrt(x))2 does not equal 49+x.

So, sometimes squaring the whole thing is the same as squaring the parts, but not generally.

I think the confusion arrives from a2 = b2 does not mean a = b


I think I explained that quite clearly, which is why you have to check the solutions you find at the end of a sequence of deductions like the one we are dealing with. But a = b does imply a2 = b2, which is what Cornstalks seems to be denying. His insistence on the distinction between squaring the sides of the equation or squaring terms makes no sense in the particular case, because there is only one term on each side of the equal sign.

But a = b does imply a2 = b2, which is what Cornstalks seems to be denying. His insistence on the distinction between squaring the sides of the equation or squaring terms makes no sense in the particular case, because there is only one term on each side of the equal sign.

I'm saying what you're really doing when you square both sides is multiply both sides by themselves. So when you have the equation:

-7 = ?x

You can't just square the terms and come up with

49 = x

You have to square both sides, which is effectively multiplying each side by itself:

(-7)2 = (?x)2

Or, another way of writing this is:

49 = ?x?x

If you don't square the sides you skip a step and assume that ?x?x = ?(xx) = x, which isn't generally true.

That's what I'm arguing.

[size=2][ I was ninja'd 71 times before I stopped counting a long time ago ] [ f.k.a. MikeTacular ] [ My Blog ] [ SWFer: Gaplessly looped MP3s in your Flash games ]

sqrt(x)^2 = x is true in general, even in the case of negative or imaginary numbers. What is the difference between squaring the "side" sqrt(x) and squaring the "term" sqrt(x)?

I'm a graduate student in math and this thread has been more convoluted than most of my textbooks - why is everyone making such a simple thing so difficult? Alvaro has the right of it.

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