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?

#Algebra

Easy Math Editor

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`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

Please solve the problem above and let me understand the mistake which I'm commiting.

Shivang Gupta - 4 years, 5 months ago

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 - 4 years, 5 months ago

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 years, 5 months ago

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 years, 5 months ago

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 years, 5 months ago

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 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$