Do derivatives introduce loss of solutions?

[Abraham]

For example, if I want to show that there is no real # solution to

x² + 24x² = -1

is it correct to show that

d²/dx²( x⁴ + 24x² ) = d²/dx²(-1)

---> 12x²+48 = 0

And since x^2 is >0 or =0, 12x²+48 ---> 0 + 48 \neq 0

Therefore, there is no real number solution to x² + 24x² = -1

Is this proper logic? Or does taking the derivatives change the validity(?) of the solution or whatever the proper terminology is?

 

[Landau]

Consider the equation x=1. Taking derivatives yield 1=0.

 

[JSuarez]

You are confusing equations with functions. The equation:

x⁴ + 24x² + 1 = 0

Asks if there is some point (in this case, a real one) such that the above equality is true, not that it's true for all points.

But when you write:

d²/dx²( x⁴ + 24x² + 1) = d²/dx²(0)

You are stating that the second derivative of the function

f(x) = x⁴ + 24x² + 1

is identically 0. This is the basic logical flaw in your argument.

(You may answer that you meant to say that they are equal at some point, but the second part of your argument contradicts this.)

 

[Abraham]

Ok, I get it now. Thanks for the for the quick replies Landau and JSuarez.