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
1 year, 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 $$x = 1 \not \Leftrightarrow x^2 = 1$$.

Staff - 1 year, 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.

- 1 year, 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.

Staff - 1 year, 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.

- 1 year, 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.

Staff - 1 year, 5 months ago

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

- 1 year, 5 months ago