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

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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

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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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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

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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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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

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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)'.

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

sqrt(25 + 25x + 25)

Edited by noatom

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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.

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