Simple manipulation of x^2 - 1 = 0, getting x=0 as a solution.

[malignant]

This isn't homework or anything, just something I tried and it doesn't seem to be consistent.

I randomly just multiplied x^2 - 1 = 0 by x instead of factoring and solving:
x^3 - x = 0. Then I factored out an x and got x(x^2 - 1) = 0.
Then I divided by x^2 - 1 and got x = 0 which isn't a solution?

I know I can just divide by x in x(x+1)(x-1) = 0 but why can't I divide by the other two factors?
what am I doing that's algebraically illegal?

 

[micromass]

What you did is correct, but problem is that your operations are irreversible.

What do I mean with that? When manipulating equations, you have almost always have an $\Rightarrow$. This means that the left-hand side implies the right hand side.

For example, $x=y~~\Rightarrow ~~ x+3 = y+3$. Indeed: if $x=y$, then I can add $3$ at both sides.
Other examples
x=y~~ \Rightarrow ~~ x^2 = y^2
x=y ~~ \Rightarrow ~~ 0x=0y

Some operations are reversible. Which means that the right-hand side also implies the left-hand side. This is a very special property. Students often think that this is always true, but that is not the case. A reversible property can be written as $\Leftrightarrow$.
For example $x=y~~ \Leftrightarrow ~~x+3 = y+3$. So adding $3$ to both sides is reversible. Which means: if I ever encounter the situation x+3=y+3, then I can always conclude that x=y.

However,
x=y~~ \Leftrightarrow ~~ x^2 = y^2
is not true. For example, $1^2 = (-1)^2$, but we can not deduce that $1=-1$.

x=y ~~ \Leftrightarrow ~~ 0x=0y
is also not true. We have that $0*1 = 0*2$, but we can not deduce that $1=2$.

Now, when solving an equation, you try to do this by manipulating your equation. The problem is however that not all manipulations are reversible.
An easy case is the following: solve 2x + 4 = 0
We can do the following:

\begin{eqnarray*}
2 x + 4 = 0
& \Leftrightarrow & 2x = -4\\
& \Leftrightarrow & x = -2
\end{eqnarray*}

What we proved now is actually two things. We have proved the normal direction: IF 2x+4=0 THEN x=-2. We have also proven the backward direction: IF x=-2 THEN 2x+4 = 0.

However, this backward direction is very often missing when solving equations. For example, let's solve $\sqrt{x^2+x+3} = x$

\begin{eqnarray*}
\sqrt{x^2+x+3} = x
& \Rightarrow & x^2 + x+ 3 = x^2
& \Leftrightarrow & x = -3
\end{eqnarray*}

In the second step, we have just squared both sides of the equation. But we know that this is not reversible!
Indeed, what we have proven now is that IF $\sqrt{x^2 + x + 3}=x$, then $x=-3$. But we can make no statement about the converse because the implications are not reversible. Indeed, we see that $x=-3$ is not a solution of the equation.

Now, in your example, you do the following
x^2 - 1 = 0~~\Rightarrow~~ x(x^2 - 1)=0

This is certainly valid. But it is not reversible. So if we are given $x(x^2 -1)=0$, then we cannot deduce that $x^2 - 1 = 0$.

So in your example, you have proven that IF $x^2 - 1 =0$, then either $x=0$ or $x=1$ or $x=-1$. This is certainly true. But the converse is not true.

 

[Mark44]

This isn't homework or anything, just something I tried and it doesn't seem to be consistent.

I randomly just multiplied x^2 - 1 = 0 by x instead of factoring and solving:
x^3 - x = 0. Then I factored out an x and got x(x^2 - 1) = 0.
Then I divided by x^2 - 1 and got x = 0 which isn't a solution?

I know I can just divide by x in x(x+1)(x-1) = 0 but why can't I divide by the other two factors?
what am I doing that's algebraically illegal?

A shorter explanation involves the solution set for each step.

Equation Solution set
1. x² - 1 = 0 {1, -1}
2. x(x² - 1) = 0 {0, 1, -1}
3. x = 0 {0}

Going from 1 to 2, you increased the solution set.
Going from 2 to 3, you decreased the solution set.

As micromass explained, your steps are not reversible, meaning that the equations are not equivalent to each other. It's only when equations are equivalent that they have the same solution set.

 

[the_wolfman]

You are dividing by zero.

In the first line you define x^2-1 =0

Thus when you divide x(x^2-1) by (x^2-1) you are dividing by zero.

 

[Mark44]

You are dividing by zero.

Not necessarily.

In the first line you define x^2-1 =0

Thus when you divide x(x^2-1) by (x^2-1) you are dividing by zero.

Only if x happens to be either 1 or -1. For all other values of x, the division is valid.

 

[the_wolfman]

Not necessarily.

Yes necessarily!

The starting equation is x^2-1 =0
Thus we know that x^2-1 =0. So dividing by x^2-1 is equivalent to dividing by zero.

Always!

Only if x happens to be either 1 or -1. For all other values of x, the division is valid.

Again from the original equation x^2-1 =0 we know that x has to be 1 or -1. So the division is never valid.

 

[Mentallic]

micromass, that was a great post, and I have but one query:

For example, let's solve $\sqrt{x^2+x+3} = x$

\begin{eqnarray*}
\sqrt{x^2+x+3} = x
& \Rightarrow & x^2 + x+ 3 = x^2
& \Leftrightarrow & x = -3
\end{eqnarray*}

In the second step, we have just squared both sides of the equation. But we know that this is not reversible!
Indeed, what we have proven now is that IF $\sqrt{x^2 + x + 3}=x$, then $x=-3$. But we can make no statement about the converse because the implications are not reversible. Indeed, we see that $x=-3$ is not a solution of the equation.

While it makes sense to say that if we have an equation and the reduction steps that arrive to the solution set are reversible, then the solution set satisfies the original equation, but now that the steps are irreversible, what does it mean to say that

"IF $\sqrt{x^2 + x + 3}=x$, then $x=-3$" ?

Yes necessarily!

The starting equation is x^2-1 =0
Thus we know that x^2-1 =0. So dividing by x^2-1 is equivalent to dividing by zero.

Always!

Yes, but all that happens is that we lose its solutions. Since solving x^2-1=0 gives us x=1,-1, then dividing by x^2-1 means that the equation we're left with has to exclude x=1,-1 from its domain.

If we instead divided by just x+1, then we're left with x-1=0 and we still have one of the solutions of the original equation left.

 

[pwsnafu]

You are dividing by zero.

In the first line you define x^2-1 =0

Thus when you divide x(x^2-1) by (x^2-1) you are dividing by zero.

Irrelevant. Even if OP didn't divide $x(x^2-1)$ by $x^2 - 1$, he would still have $x = 0$ as solution (which isn't a solution in the original equation).

 

[Mark44]

You are dividing by zero.

Not necessarily.

In the first line you define x^2-1 =0

Thus when you divide x(x^2-1) by (x^2-1) you are dividing by zero.

Only if x happens to be either 1 or -1. For all other values of x, the division is valid.

Yes necessarily!

The starting equation is x^2-1 =0
Thus we know that x^2-1 =0. So dividing by x^2-1 is equivalent to dividing by zero.

Always!

Again from the original equation x^2-1 =0 we know that x has to be 1 or -1. So the division is never valid.

When you say "always", the implication is that you mean "for all x." The expression x² - 1 is not identically equal to zero. As I pointed out before, this expression is equal to zero only if x = 1 or x = -1. For all other values of x, it is legitimate to divide by x² - 1.

In this case, the net effect is that by adding and eliminating factors on the left side, we increase or decrease the solution set.

 

[pwsnafu]

but now that the steps are irreversible, what does it mean to say that

"IF $\sqrt{x^2 + x + 3}=x$, then $x=-3$" ?

Exactly what it says on the tin

If I have $A \implies B$, then when A is true, so is B.
What happens when A is false? Well we get no information about B.

What happens when A is always false? We get no information about B what so ever!
Nonetheless, we do this in everyday language: "if I had caught the bus to work, I would have been late". This is called a counterafactual.

 

[micromass]

micromass, that was a great post, and I have but one query:



While it makes sense to say that if we have an equation and the reduction steps that arrive to the solution set are reversible, then the solution set satisfies the original equation, but now that the steps are irreversible, what does it mean to say that

"IF $\sqrt{x^2 + x + 3}=x$, then $x=-3$" ?

It means that the only possible solution to the equation is -3. However, it is not guaranteed that it actually is an solution. We just know that if there is a solution, then it must be -3. However, it may happen that there are no solutions! To check whether -3 is a solution, we need to substitute it in the equation.

 

[rcgldr]

I randomly just multiplied x^2 - 1 = 0 by x instead of factoring and solving: x^3 - x = 0. Then ... got x = 0 which isn't a solution?

x = 0 is one of 3 roots for the cubic equation you created, so there's no conflict.

For a more general example, you could choose any 3rd root r, and multiply x^2 - 1 = 0 by (x-r), to produce a cubic equation that includes x = r as a root, in addition to x = 1, and x = -1.

 

[the_wolfman]

When you say "always", the implication is that you mean "for all x." The expression x2 - 1 is not identically equal to zero.

No! I don't mean for all x.

I randomly just multiplied x^2 - 1 = 0 by x

The starting equation and the title of this post is x^2-1=0. Once you define x^2-1 to be zero it is zero. End of story. If you then try to divide by x^2-1 you're going to have a bad day. And this is why the original poster wrongly gets the x=0 is a root of x^2-1=0.

Once you define x^2-1 =0 you can infer that x=1 or x=-1.

 

[pwsnafu]

If you then try to divide by x^2-1 you're going to have a bad day. And this is why the original poster wrongly gets the x=0 is a root of x^2-1=0.

No. The division is not the cause of the problem.
$x^2 - 1 = 0 \implies x(x^2-1) = 0$
is valid.
$x(x^2-1)=0 \implies x^2-1 = 0$
is not.

 

[DeeAytch]

Not necessarily.
Only if x happens to be either 1 or -1. For all other values of x, the division is valid.
When you say "always", the implication is that you mean "for all x." The expression x² - 1 is not identically equal to zero. As I pointed out before, this expression is equal to zero only if x = 1 or x = -1. For all other values of x, it is legitimate to divide by x² - 1.

All x in the set of solutions, not in the set of real numbers. Then, all x means x=1 and x=-1. I'm sure that this is what he meant.

If we are changing the solution set of a polynomial, then we are changing the characteristic property of that polynomial. Hence, we would have a different polynomial.

 

[DeeAytch]

No. The division is not the cause of the problem.
$x^2 - 1 = 0 \implies x(x^2-1) = 0$
is valid.
$x(x^2-1)=0 \implies x^2-1 = 0$
is not.

By hypothesis $x^2-1=0$. We don't need that last implies.

The division is the cause of the problem.

$x(x^2-1)=0 \implies x = 0$

Is invalid. Really, it implies that $x=0$ or $x=1$ or $x=-1$. But, it doesn't equal zero because that would not satisfy the condition in the hypothesis.

 

[pwsnafu]

By hypothesis $x^2-1=0$. We don't need that last implies.

The division is the cause of the problem.

But no division is necessary. You still get the same problem.

We start with
$x^2 - 1 = 0$
$\implies x(x^2-1) = 0$.
Now from the fact that $\mathbb{R}[X]$ has no zero divisors, we know
$\implies x = 0 \vee x^2-1 =0$
and now I have the same problem. But I have not done any division at all.

Re-read micromass's post. In symbols it goes:
$x=1 \vee x=-1 \iff x^2-1 = 0 \implies x(x^2-1) = 0 \iff x=1 \vee x=0 \vee x=-1$
That middle $\implies$ cannot be reversed.

PS: Khrisstian you seem to not have realized that the hypothesis has swapped is now the cubic, not the quadratic. You can't assume $x^2 - 1 = 0$ is true any more because that is what you are trying to prove now.

 

[DeeAytch]

I don't think there was a hypothesis imposter. The hypothesis was that x^2-1=0. Why would that change?

Starting with that hypothesis you know that x is 1 or -1. If you multiplied both sides of the equation by anything at all, then the equation would remain true. This is precisely because x is 1 or -1. All other derived solutions must be in the set {1,-1} to be actual solutions for x.

 

[DeeAytch]

x has graduated to playing a role in a system of equations once you start multiplying both sides of the first equation. Equation 1 still needs to be true.

 

[pwsnafu]

I don't think there was a hypothesis imposter. The hypothesis was that x^2-1=0. Why would that change?

Because we are interested in the opposite direction. That means you read backwards: if you structured the proof vertically you now read bottom up. And you are trying to prove what you assumed at the start.

You seem to not to understand the problem itself. So I'll break it down. The OP started with an axiom, let's call it Statement A. Statement A says "$x^2-1=0$". He then uses Statement A to prove Statement B. Statement B says "$x(x^2-1) = 0$".

So we have $A \implies B$. Up to here no problem. But two things happen. First, he realizes that Statement A is equivalent to another Statement A'. A' says $x=\pm1$.
Secondly, Statement B is also equivalent...to Statement B', which says (you guessed it!) $x = 0, \pm 1$.

And now OP asks us: why does B' contain 0, even though A' does not.

The reason? A is not equivalent to B. Equivalent means $A \implies B$ and $B \implies A$. We know the first implication is correct, we already did that one! Nobody is complaining about that one. It's the second implication that causes the problem. OP is unconsciously assuming that A "should be" equivalent to B, because that's what he has previous experience with.

But in this example $B \not\Longrightarrow A$. This is why I keep saying the hypothesis swapped. We assume B, and look at whether A can be proved from B. You can't assume what you are trying to prove.

Edit: PS

You might ask, why should we care? Well, it's actually the definition of "a solution to the equation..."
To see this consider

-3 is the solution to 2x +4 = -2

in symbols we get

$x=-3 \implies 2x+4=-2$

Notice the direction of the arrow. But when you solve the equation you do
$2x+4 = -2$
$\implies 2x = -6$
$\implies x = -3$
The arrows are the wrong way!

We have not "solved" the problem, actually what we did was show
$2x+4 = -2 \implies x=-3$, or in words, "if there is a solution, that solution is -3".
-3 is not (yet) the solution, it's a candidate to the solution. To complete the problem, "we check our answer":
$x=-3 \implies 2x = -6 \implies 2x+4 =-2$
and our arrows are all in the right direction. The "plug the answer back" is what makes the proof work.

 

[DeeAytch]

We are showing two sets (the solution sets) are not equal. It doesn't matter where we start.

I also addresses the issue of why the x he picked didn't satisfy the system of equations he constructed. This had the hypothesis of x^2-1=0.

 

[the_wolfman]

And now OP asks us: why does B' contain 0, even though A' does not.

Actually OP asks us

I know I can just divide by x in x(x+1)(x-1) = 0 but why can't I divide by the other two factors?
what am I doing that's algebraically illegal?

And the reason is that in order for the hypothesis x^2-1=0 to be true x=1 or x=-1. Thus when dividing by the product (x+1)(x-1) he is dividing by zero. Dividing by zero is algebraically illegal. No other step in his derivation is illegal.

While the step where he multiplies both sides of the equation by x does introduce the root x=0 to the system of equations. It is still algebraically legal. x^3-x=0 is true if x^2-1=0. Its only when he divides by x^2-1=0 does he get a answer (x=0) that is odds with the original equations.

In fact if you skip multiplying by x and just divide both sides of the equation by x^2-1 you will still get a nonsensical answer.
x^2-1=0
\frac{x^2-1}{x^2-1}=\frac{0}{x^2-1}
1=0

The reason why this does not work is that you are dividing by zero.

 

[pwsnafu]

We are showing two sets (the solution sets) are not equal.

And I'm explaining why, because the original poster was manipulating equations.

It doesn't matter where we start.

All mathematics are sequences of logical arguments. All mathematics by definition depends on where you start.

And two sets, A and B, are equal (A=B) if and only if $A \subset B$ and $B \subset A$. Everything I said applies.

I also addresses the issue of why the x he picked didn't satisfy the system of equations he constructed. This had the hypothesis of x^2-1=0.

Serious question. Do you understand what micromass and I are saying?

 

[pwsnafu]

Cross posted

While the step where he multiplies both sides of the equation by x does introduce the root x=0 to the system of equations. It is still algebraically legal. x^3-x=0 is true if x^2-1=0.

Yes! $x^2-1=0 \implies x(x^2-1)=0$.
I even said so: $A \implies B$
But it's not reversible. Did you read micromass's post?

Its only when he divides by x^2-1=0 does he get a answer (x=0) that is odds with the original equations.

No he also made a more fundamental error: assuming reversibility.

In fact if you skip multiplying by x and just divide both sides of the equation by x^2-1 you will still get a nonsensical answer.
x^2-1=0
\frac{x^2-1}{x^2-1}=\frac{0}{x^2-1}
1=0

The reason why this does not work is that you are dividing by zero.

Yeah, I know...and...?

Edit: just to clear something up, if you think I agree with Mark's posts, then no. I'm not saying the division is valid. But you'll still get an error even if you don't divide. That's the point. You are claiming the division is the only error he's made. That's wrong.

 

[DeeAytch]

It doesn't matter where we start since inequality is reflexive. Showing a set A does not equal a set B is just as good as showing the set B does not equal the set A.

We were done when we found both sets and found the counter example, 0, in the second equation's solution set.

So, we can conclude that (1) x(x^2-1)=0 and (2) x^2-1=0 are different polynomials with different roots. (:

Do note that because (2) is a factor of (1), we can see that (2)'s solutions form a subset less than or equal to (1)'s solutions. More generally, whatever polynomial you multiply (2) by with result in a polynomial whose solution set contains the solutions of (2). This is precisely why we factor polynomials to find their roots. (or wolfram it)

 

[pwsnafu]

It doesn't matter where we start since inequality is reflexive.

I'm assuming you mean equality not inequality.

Showing a set A does not equal a set B is just as good as showing the set B does not equal the set A.

In order to show $A \neq B$ you need either show $A \not\subset B$ or $B \not\subset A$. (Or both, but only one is necessary.)

In this situation $A \subset B$ but $B \not\subset A$. The original post is making the assumption they should be equal. We are explaining why that is invalid.

We were done when we found both sets and found the counter example, 0, in the second equation's solution set.

So, we can conclude that (1) x(x^2-1)=0 and (2) x^2-1=0 are different polynomials with different roots. (:

Do note that because (2) is a factor of (1), we can see that (2)'s solutions form a subset less than or equal to (1)'s solutions. More generally, whatever polynomial you multiply (2) by with result in a polynomial whose solution set contains the solutions of (2). This is precisely why we factor polynomials to find their roots. (or wolfram it)

You're not disagreeing with what I'm saying.

The problem is the original poster did make the assumption that they "should have" the same roots (i.e. that his manipulations were reversible) and ended up in confusion.

 

[malignant]

Now, in your example, you do the following
x^2 - 1 = 0~~\Rightarrow~~ x(x^2 - 1)=0

This is certainly valid. But it is not reversible. So if we are given $x(x^2 -1)=0$, then we cannot deduce that $x^2 - 1 = 0$.

So in your example, you have proven that IF $x^2 - 1 =0$, then either $x=0$ or $x=1$ or $x=-1$. This is certainly true. But the converse is not true.

This is really weird to me when I think of the solution sets. Intuitively, I figured {0,-1,1} could imply {-1,1} since the latter is a subset, and that {-1,1} couldn't imply {0,-1,1}. But it's legal to expand the solution set by multiplying so I guess that makes sense, but what about dividing to reduce the solution sets?

What if I started with x(x^2 - 1) = 0, that can imply x^2 - 1 = 0? Then is it not reversible the other way around in this case?

 

[pwsnafu]

This is really weird to me when I think of the solution sets. Intuitively, I figured {0,-1,1} could imply {-1,1} since the latter is a subset, and that {-1,1} couldn't imply {0,-1,1}.

Your intuition was the other way around.

Edit: to elaborate, the solution sets represent OR not AND.
{-1,1} means "x = -1 OR x = 1".
{-1, 0, 1} means "x = -1 OR x = 1 OR x = 0".
So the former implies the latter.

But it's legal to expand the solution set by multiplying so I guess that makes sense, but what about dividing to reduce the solution sets?

What if I started with x(x^2 - 1) = 0, that can imply x^2 - 1 = 0?

No. x(x^2-1) = 0 implies (x^2-1=0 OR x=0). I put parenthesis there for emphasis. You cannot just throw away a possible solution.

Then is it not reversible the other way around in this case?

$x^2 - 1 = 0 \implies x(x^2-1) = 0$ is indeed valid.

 

[Crazy Horse 11]

Yes necessarily!

The starting equation is x^2-1 =0
Thus we know that x^2-1 =0. So dividing by x^2-1 is equivalent to dividing by zero.

Always!



Again from the original equation x^2-1 =0 we know that x has to be 1 or -1. So the division is never valid.

do you know what has just been proven by x^2 - 1 = 0? That 0/0=1. see?

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

P.j.S .

 

[micromass]

do you know what has just been proven by x^2 - 1 = 0? That 0/0=1. see?

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

P.j.S .

The expression 0/0 is undefined. The expression

\frac{x^2 -1}{x^2 - 1}

is therefore undefined for all $x$ which make the denominator $0$.

 

[Crazy Horse 11]

The expression 0/0 is undefined. The expression

\frac{x^2 -1}{x^2 - 1}

is therefore undefined for all $x$ which make the denominator $0$.

0/0=1 may be undefined but (x^2 - 1) / (x^2 - 1) when x=(-1, 1) proves it.

P.j.S .

 

[micromass]

0/0=1 may be undefined but (x^2 - 1) / (x^2 - 1) when x=(-1, 1) proves it.

P.j.S .

If $x=1$, then

\frac{x^2-1}{x^2 -1}

is undefined. Same with $x=-1$.

 

[Crazy Horse 11]

If $x=1$, then

\frac{x^2-1}{x^2 -1}

is undefined. Same with $x=-1$.

(x^2 - 1) / (x^2 - 1) = 1 is that true or false?

 

[micromass]

(x^2 - 1) / (x^2 - 1) = 1 is that true or false?

Depends on the value of $x$.

 

[Crazy Horse 11]

Depends on the value of $x$.

Something divided by itself always = 1. x^2 - 1 is no different.

Therefore for example when x=1 then 0/0=1:

(x^2 - 1) / (x^2 - 1) = 1. This may be undefined but it is proved true by the fundamental mentioned above.

Prove that 0 / 0 = 1 is not true.

 

[micromass]

Something divided by itself always = 1.

Not true.

x^2 - 1 is no different.

Therefore for example when x=1 then 0/0=1:

(x^2 - 1) / (x^2 - 1) = 1. This may be undefined but it is proved true by the fundamental mentioned above.

Prove that 0 / 0 = 1 is not true.

It's essentially a definition. The division "x/y" is defined for all real (and complex numbers) except for y=0. We have left "x/0" undefined by definition.

The reason is simple. What would we define it as?? You could argue that

\frac{x^2 - 1}{x^2 - 1}=1

for all $x$ except $1$ and $-1$. So you could say that it makes sense to let this equality hold also for $1$ and $-1$. That way, we could define $0/0=1$.
The reason we don't do this, is because we could also argue that for $x\notin \{1,-1\}$ that

\frac{2(x^2-1)}{x^2-1}=2

So we want this to hold also for $1$ and $-1$. But that forces $0/0=2$.

In the same way, we could make a perfectly fine argument that $0/0=r$ for any number r! So there is no definition of $0/0$ which unambiguously extends the laws of arithmetic. This is why we define $0/0$ to be undefined.

 

[Crazy Horse 11]

Not true.



It's essentially a definition. The division "x/y" is defined for all real (and complex numbers) except for y=0. We have left "x/0" undefined by definition.

The reason is simple. What would we define it as?? You could argue that

\frac{x^2 - 1}{x^2 - 1}=1

for all $x$ except $1$ and $-1$. So you could say that it makes sense to let this equality hold also for $1$ and $-1$. That way, we could define $0/0=1$.
The reason we don't do this, is because we could also argue that for $x\notin \{1,-1\}$ that

\frac{2(x^2-1)}{x^2-1}=2

So we want this to hold also for $1$ and $-1$. But that forces $0/0=2$.

In the same way, we could make a perfectly fine argument that $0/0=r$ for any number r! So there is no definition of $0/0$ which unambiguously extends the laws of arithmetic. This is why we define $0/0$ to be undefined.

What is your answer for 2(x^2 - 1) / (x^2 - 1) = ?

 

[micromass]

What is your answer for 2(x^2 - 1) / (x^2 - 1) = ?

It is undefined for $x=1$ and $x=-1$. It is $2$ otherwise.

 

[Crazy Horse 11]

It is undefined for $x=1$ and $x=-1$. It is $2$ otherwise.

2(x^2 - 1) / (x^2 - 1) = x^2 - 1 if x=5 then the answer is 24. If x=1 then the answer = 0. Please show how you got the answer of 2.

 

[Fredrik]

2(x^2 - 1) / (x^2 - 1) = x^2 - 1 if x=5 then the answer is 24. If x=1 then the answer = 0. Please show how you got the answer of 2.

For all real numbers x except 0, we have x/x=1. This implies that for all real numbers x except 1 and -1, we have
$$\frac{2(x^2-1)}{(x^2-1)}=2\cdot 1=2.$$

 

[micromass]

2(x^2 - 1) / (x^2 - 1) = x^2 - 1 if x=5 then the answer is 24. If x=1 then the answer = 0.

Wut?

 

[Fredrik]

So you are trying to teach me that 2(3)/3 = 2*1?

Yes, because what you said in post #39 implies that it's not.

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

This is wrong, but now at least I understand what you're doing wrong. $2(x^2-1)$ doesn't mean what you think it does. It's just the number 2 times the result of $x^2-1$. It doesn't mean the same thing as $(x^2-1)^2$.

 

[Crazy Horse 11]

Yes, because what you said in post #39 implies that it's not.


This is wrong, but now at least I understand what you're doing wrong. $2(x^2-1)$ doesn't mean what you think it does. It's just the number 2 times the result of $x^2-1$. It doesn't mean the same thing as $(x^2-1)^2$.

Thank-you for pointing out my oversight.