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#1 2012-12-20 12:01:58

Frublox
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Square root problem

Got this problem on a test:
____
12 + √2x-1 = 4

Okay, so then I subtract 12 from both sides:
____
√2x-1 = -8
2x - 1 = 64   I then square both sides...
2x = 65        Add 1 to each side
x = 65/2       Divide both sides by two.

However, that answer doesn't work. Are there simply no solutions, then?

#2 2012-12-20 13:21:46

bobbym

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Re: Square root problem

Hi Frublox;

That is correct, there are no solutions. Not every equation is a true statement. You always check the validity of your answer by plugging into the original equation.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#3 2012-12-20 19:45:43

bob bundy
Moderator

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Re: Square root problem

hi Frublox

That answer is the solution to

-12 + √(2x-1) = -4

When you square you cannot help but introduce 'solutions' to this alternative equation as well so you were right to check the value you had.

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

#4 2012-12-20 20:25:17

anonimnystefy
Real Member

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Re: Square root problem

Even when you got √2x-1 = -8 you could've said that there are no solutions, because a square root of a positive number is always positive...

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#5 2012-12-20 20:31:14

bobbym

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Re: Square root problem

Hi;

Yes, it always takes the principal value.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.