Question about the characteristics of equations.

[Curd]

I may just be overlooking something here but...


4-(1/x)-(2/x^2)=0

converts to

(4x^2)-(1x)-2=0

if you multiply the first one by x^2

BUT! they do not graph the same.

BUT! to make it easy to work in the quadratic equation you have to multiply the original by x^2


what gives?

are there characteristics to equations that are not addressed by the distribution and order of operations properties, or did i just miss something?

the answer book said this method was correct.



and this is not for homework BTW. I already graduated university and am in the process of going over all the math i took back in school, college algebra especially as i took it in a 20 day may course. I want to go past the calculus book this time so I'm taking it slow and asking all the questions i need to.

 

[Stephen Tashi]

The operations used in solving equations do change the functions involved in the equations so you should expect the associated graphs to change. What we demand of operations that solve equations is that they do not change the roots of the equation. If an equation of the form f(x) = 0 is operated upon and we get a new equation g(x) = 0 then f(x) and g(x) may have different graphs, but their graphs should intersect the x-axis at the same places.

Incidentally, when you multiply that equation by x^2 , you are supposed make a note (at least a mental note) that x^2 is not zero. Multiplying both sides of an equation by a non-zero number doesn't change the roots, but multiplying both sides by zero might.

 

[Hurkyl]

Incidentally, when you multiply that equation by x^2 , you are supposed make a note (at least a mental note) that x^2 is not zero. Multiplying both sides of an equation by a non-zero number doesn't change the roots, but multiplying both sides by zero might.

To be fair, this case is slightly different. That x is nonzero is implicit in the first equation (because it divides by x). However, such a note is still a good to make, because this constraint is no longer implicit in the second equation, so it could be easy to forget.

 

[Mark44]

I may just be overlooking something here but...


4-(1/x)-(2/x^2)=0

converts to

(4x^2)-(1x)-2=0

if you multiply the first one by x^2

BUT! they do not graph the same.

Yes they do!

A graph of the first equation consists of two points on the number line. A graph of the second equation consists of the same two points.

Now if you graph f(x) = 4 - 1/x - 2/x² and g(x) = 4x² - x -2, the graphs are of course going to be different, but they will both cross the x-axis at the same two points, and that's what you ostensibly cared about when you solved the equation 4x² - x -2 = 0.

All that you were really concerned with were the solutions to this equation.