two solutions

[chrislav]

In the book : CORE MATHS FOR A- LEVEL,on page 41 I found the folllowing:

1st solution

$\displaystyle x^2-5x=0$

$\displaystyle x(x-5)=0$

hence x=0 or x=5

2nd solution

$\displaystyle x^2-5x=0$
$\displaystyle x-5=0$ dividing by x

hence the solution x=0 has been lost

is the above correct?

 

[Dr.Peterson]

is the above correct?

It's correct, as long as you understand what it is saying: that the second method is wrong (unless you recognize the issue and separately check whether x=0 is a solution).

You should never divide by the variable (or an expression containing the variable) unless either you know that what you divide by is not zero, or you consider that possibility separately. When you are tempted to divide, you can usually replace that step with factoring, as in the first method,.

 

[Dr.Peterson]

I should have added that dividing by x is a far greater temptation when the equation is written as $x^2=5x$, where dividing by x is the obvious thing to do, and factoring is very unnatural.

 

[chrislav]

In the book : CORE MATHS FOR A- LEVEL,on page 41 I found the folllowing:

1st solution

$\displaystyle x^2-5x=0$

$\displaystyle x(x-5)=0$

hence x=0 or x=5

2nd solution

$\displaystyle x^2-5x=0$
$\displaystyle x-5=0$ dividing by x

hence the solution x=0 has been lost

is the above correct?

 

[chrislav]

so by the second solution the solution x=0 is lost although we dividing by x different than zero
and we have to consider whether x=0 is a solution

 

[Dr.Peterson]

although we dividing by x different than zero

I suppose you mean that in dividing by x, you are implicitly assuming that it is non-zero. That's right. It should be stated explicitly.

we have to consider whether x=0 is a solution

Correct. If you do that division, you must check that possibility.

So you might show your work something like this:

We are solving $x^2-5x=0$​

​

We see by inspection that $x=0$ is a solution.​

​

Now, assuming that $x\ne0$, we can divide by x to obtain $x=5$.​

 

[blamocur]

$x = 2.5 \pm \sqrt{2.5^2 - 0}$

 

[chrislav]

I suppose you mean that in dividing by x, you are implicitly assuming that it is non-zero. That's right. It should be stated explicitly.


Correct. If you do that division, you must check that possibility.

So you might show your work something like this:

We are solving $x^2-5x=0$​

​

We see by inspection that $x=0$ is a solution.​

​

Now, assuming that $x\ne0$, we can divide by x to obtain [imat=5[/imath].h]x​

​

Now, assuming that $x\ne0$, we can divide by x to obtain x=5​

So we actualy have proved :$x\ne0$ implies x=5​

But isnt that equivalent to x=0 or x=5 accordingic to logic​

​

¬p implies q this is equivalent to p or q​

Hence dividing by x,different to zero we do not lose a solution as the book claims​

 

[Dr.Peterson]

But isn't that equivalent to x=0 or x=5 according to logic

Yes, that was my point. Was that not clear?

This is a way to modify the invalid second method into a method that gives the same answer as the first. (Any valid method has to give the same answer!)

But I wasn't trying to invoke that sort of logic, but just common sense: Having seen that one solution is x=0, we can look for another solution under the assumption that x is not 0. So now we have found two solutions.

Hence dividing by x,different to zero we do not lose a solution as the book claims

We do lose a solution if we do only what the book showed in that second method. We had to fix that method to make it valid.

Of course, the factoring method is more straightforward.

 

[chrislav]

how can one invoke logic by using common sense
what is common sense anyway

 

[Dr.Peterson]

Common sense is unstated (and often unconscious) logic: "sound practical judgment that is independent of specialized knowledge, training, or the like; normal native intelligence". I didn't call upon logical symbols or theorems; that's all I was saying, that you don't need to use formal logic to understand the idea.

how can one invoke logic by using common sense

But that's the opposite of what I said, which was "I wasn't trying to invoke that sort of logic".