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An amazing paradox

Consider the equation \[ x^2 + x +1 = 0 \tag {1} \]

Factorizing it, we get \[ (x + \sqrt x + 1)(x - \sqrt x + 1) = 0 , \] or \[ x + 1 = \pm \sqrt x \tag {2} \]

Substituting \((2) \) into \((1) \), we get \( \sqrt x = 0 \) as a root of equation \((1) \). And the first equation does not satisfy \(x=0 \), so how is this possible?

Note by Shivang Gupta
5 months ago

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Remember that as you manipulate equations, you may introduce extraneous roots. Do all your equations have double-sided implication signs? EG \( x = 1 \not \Leftrightarrow x^2 = 1 \). Calvin Lin Staff · 5 months ago

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@Calvin Lin But I've not square my function in any step to introduce any extrenious roots. I have just implied my observation of function by substituting in it. Shivang Gupta · 4 months, 4 weeks ago

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@Shivang Gupta Apart from squaring, when you add or subtract equations, you may also introduce extraneous roots.

For a simple example, if we wanted to solve \( x = 0 , y = 0 \), we cannot add them up and say that oh \( x+y = 0 \), and observe that \( (x,y) = (-1, 1) \) is a solution!

This is why it's important to place implication signs everwhere, to the extent that they are valid. IE

\[ x=0, y = 0 \Rightarrow x+y = 0 \Rightarrow (x,y) = (-1, 1) \]

which tells you that the final condition could have been an extraneous solution. Calvin Lin Staff · 4 months, 4 weeks ago

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@Calvin Lin In the above equation I've used only one variable and I am unable to understand that if 2 things are coming out to be equal then why they can't be substituted. In the example you mentioned you took 2 variable but I used only one variable. If you do not agree with me then please indicate the step which I did wrong with proper reasoning. Shivang Gupta · 4 months, 3 weeks ago

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@Shivang Gupta As I said, the step where you went wrong is "Apart from squaring, when you add or subtract equations, you may also introduce extraneous roots.". The reason is that "If \(x\) is a solution to equation 1 and equation 2 \( \Rightarrow \) then \(x\) is a solution to equation 1 + equation 2", but not vice versa.

Here's a ridiculous counter example to make it clearer. Suppose \( x= 1 \). Adding \( -x = - 1 \) to both sides, we get \( 0 = 0 \). Hence, this equation is always true, so \(x\) is any value.

This is why it's always important to place implication signs to illustrate the extent that they are valid. IE \( f(x) = 0 , g(x) = 0 \Rightarrow f(x) + g(x) = 0 \) but not vice versa. Calvin Lin Staff · 4 months, 3 weeks ago

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Please solve the problem above and let me understand the mistake which I'm commiting. Shivang Gupta · 5 months ago

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