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Radical equation


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#1 noatom   Members   -  Reputation: 782

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Posted 12 November 2013 - 01:07 PM

3√x + 8 = 10√x + 9 

 

This is how i solved it:

 

subtract 3√x from both sides,so the result is:

 

8 = 7√x + 9

 

subtract 9 from both sides

 

-1 = 7√x

 

after that,everything gets squared:

 

1 = 49 x 

 

and then,divide everything by 49,so:

 

1/49 = x

 

HOWEVER,there must be something wrong,cause the above equation has no solutions,due to the fact that the response should be negative.

 

But mine,obviously has a solution....What am i doing wrong?



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#2 h4tt3n   Members   -  Reputation: 1051

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Posted 12 November 2013 - 01:35 PM

It can't be solved. square root x can be replaced by any number, including 1. So, the equation says 3 + 8 = 10 + 9.



#3 noatom   Members   -  Reputation: 782

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Posted 12 November 2013 - 01:38 PM

That doesn't answer my question....what i want to know,is what i did wrong in the solving of the equation.

 

If you get that in a test you can't just start replacing x with random values to prove it works/doesn't...

 

 

Remember that the the solution is: there is no solution because the answer is a negative number.

What i have there is 1/49,which is not negative,so i must've done something wrong.


Edited by noatom, 12 November 2013 - 01:49 PM.


#4 Cornstalks   Crossbones+   -  Reputation: 6989

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Posted 12 November 2013 - 01:49 PM

You didn't "square both sides". You squared each term. Which is not the same. If you square each term instead of each side, you skip a few steps and assume √(zw) = √z√w, which is in general not true.


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#5 Álvaro   Crossbones+   -  Reputation: 13311

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Posted 12 November 2013 - 02:05 PM

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.

 

Another way of saying it is that your argument is a bunch of implications: "If this equation is true, then this other equation is true, and then this other equation is true, etc." So you proved that, if there is a solution, it must be x=1/49. However, the implications don't work in reverse (in particular the squaring step can say something like "-2=2 implies 4=4", but the reciprocate is not true). So you have to verify that x=1/49 is indeed a solution, which it isn't. So you proved that there are no solutions.


Edited by Álvaro, 12 November 2013 - 02:06 PM.


#6 Cornstalks   Crossbones+   -  Reputation: 6989

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Posted 12 November 2013 - 02:22 PM

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.


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#7 noatom   Members   -  Reputation: 782

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Posted 12 November 2013 - 02:39 PM

Ok,at first i was a little bit confused by cornstalks's affirmation : " squaring a side does not equal squaring  the terms of that side".

But i finally got it,example:

 

(2x-6)^2

 

If i were to square each term,the result would be

 

2*2 = 4, x*x=x^2, -6*-6 = 36

=> 4x^2 + 36

 

But the right way(squaring a side,not terms):

 

(2x-6) * (2x-6)

2x * 2x = 4x, 2x * -6 = -12x, -6*2x = -12x, -6 * -6 = 36

 

=> 4x -12x -12x + 36

4x - 24x + 36

 

So yeah...you have to be really carefull with these things...

 

And a big note,even though Cornstalks already said it: "Doing ANYTHING to a side does not equal doing ANYTHING to the members of that side",where anything can be multiplication,division...you get the ideea..


Edited by noatom, 12 November 2013 - 02:43 PM.


#8 Álvaro   Crossbones+   -  Reputation: 13311

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Posted 12 November 2013 - 02:45 PM

 

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)'.



#9 noatom   Members   -  Reputation: 782

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Posted 12 November 2013 - 02:53 PM

em...is it true that you cannot simplify this?

 

sqrt(25 + 25x + 25)


Edited by noatom, 12 November 2013 - 02:57 PM.


#10 Álvaro   Crossbones+   -  Reputation: 13311

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Posted 12 November 2013 - 03:05 PM

It depends on what you mean by "simplify":

 

sqrt(25 + 25x + 25) = 5 * sqrt(2 + x)

 

I think that's simpler, but you might disagree.



#11 Cornstalks   Crossbones+   -  Reputation: 6989

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Posted 12 November 2013 - 03:13 PM

 

 

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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#12 Paradigm Shifter   Crossbones+   -  Reputation: 5372

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Posted 12 November 2013 - 05:16 PM

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)).


Edited by Paradigm Shifter, 12 November 2013 - 05:18 PM.

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#13 Hodgman   Moderators   -  Reputation: 30384

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Posted 12 November 2013 - 05:25 PM

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+

To9Yxbh.gif

daizCFc.gif

LsqJZHt.gif



#14 Paradigm Shifter   Crossbones+   -  Reputation: 5372

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Posted 12 November 2013 - 05:35 PM

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+


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#15 Álvaro   Crossbones+   -  Reputation: 13311

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Posted 12 November 2013 - 06:48 PM

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.

#16 Paradigm Shifter   Crossbones+   -  Reputation: 5372

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Posted 12 November 2013 - 07:01 PM

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


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#17 Hodgman   Moderators   -  Reputation: 30384

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Posted 12 November 2013 - 07:31 PM

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.



#18 Álvaro   Crossbones+   -  Reputation: 13311

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Posted 12 November 2013 - 07:33 PM

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.

#19 Cornstalks   Crossbones+   -  Reputation: 6989

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Posted 12 November 2013 - 08:32 PM

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.


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#20 Ezbez   Crossbones+   -  Reputation: 1164

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Posted 12 November 2013 - 08:56 PM

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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