two solutions

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τρικλινο

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#1 h Apr 12, 2025, 6:20 PM

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in a book:CORE MATHS for A-LEVEL ON PAGE 41 i found the following

1st solution

$x^2-5x=0$

$x(x-5)=0$

hence x=0 or x=5

2nd solution

$x^2-5x=0$

$x-5=0$ dividing by x

hence the solution x=0 has been lost

is the above correct?

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Safal

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#2 h Apr 12, 2025, 6:31 PM

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2nd Solution is basically wrong. Why? Here is the explanation.

$x(x-5)=0$ Then there are two cases either $x=0$ or $x\neq 0$ .When we are admitting the case $x=0$ we cannot divide by $0$ . So, in the case we apply divison by $x$ then $x\neq 0$ is a solid prerequisite to do so.Thus, $x-5=0$ from $x(x-5)=0$ we must take the assumption in hand that $x\neq 0$ . For example take the extension of the same problem in $\mathbb{F}_5$ then the same problem reads $x^2=0$ ,implying only one solution with optimistic repetation of root $0$ , two times that is the multiplicity of $0$ in $x^2$ . Thankfully, we are lucky enough that we are in the field of $\text{char}$ $0$ .The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of $\text{char}$ $0$ .

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#3 h Apr 13, 2025, 12:02 AM

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Safal wrote:

2nd Solution is basically wrong. Why? Here is the explanation.

$x(x-5)=0$ Then there are two cases either $x=0$ or $x\neq 0$ .When we are admitting the case $x=0$ we cannot divide by $0$ . So, in the case we apply divison by $x$ then $x\neq 0$ is a solid prerequisite to do so.Thus, $x-5=0$ from $x(x-5)=0$ we must take the assumption in hand that $x\neq 0$ . For example take the extension of the same problem in $\mathbb{F}_5$ then the same problem reads $x^2=0$ ,implying only one solution with optimistic repetation of root $0$ , two times that is the multiplicity of $0$ in $x^2$ . Thankfully, we are lucky enough that we are in the field of $\text{char}$ $0$ .The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of $\text{char}$ $0$ .

so how do we get x=0 or x=5 ,since we assumed $x\neq 0$ .

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maxamc

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#4 h Apr 13, 2025, 3:25 AM

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Move this to MSM, reported

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Safal

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#5 h Apr 13, 2025, 6:22 AM

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τρικλινο wrote:

Safal wrote:

2nd Solution is basically wrong. Why? Here is the explanation.

$x(x-5)=0$ Then there are two cases either $x=0$ or $x\neq 0$ .When we are admitting the case $x=0$ we cannot divide by $0$ . So, in the case we apply divison by $x$ then $x\neq 0$ is a solid prerequisite to do so.Thus, $x-5=0$ from $x(x-5)=0$ we must take the assumption in hand that $x\neq 0$ . For example take the extension of the same problem in $\mathbb{F}_5$ then the same problem reads $x^2=0$ ,implying only one solution with optimistic repetation of root $0$ , two times that is the multiplicity of $0$ in $x^2$ . Thankfully, we are lucky enough that we are in the field of $\text{char}$ $0$ .The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of $\text{char}$ $0$ .

so how do we get x=0 or x=5 ,since we assumed $x\neq 0$ .

If you read carefully I haven't said that we cannot get $x=0$ .The assumption whenever $x\neq 0$ we get $x=5$ else we get the case $x=0$ .I can explain you more but the fact is I cannot use argument of field theory to explain it in total details.The reason why it's actually the case lies in field theory logics.

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#6 h Apr 13, 2025, 1:51 PM

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what is field theory logic.
IS the logic that suports the development of field theory?
THIS post should not be moved to Middle School Math
Because in the 2nd solution we have the answer : x different than zero this implies x=5
And according to logic this is equivelant to x=0 or x=5.Hence no solution is lost as the book claims
There for it should be removed back to at least college algebra although i doupt if even there anyone knew of that solution
WEmake use of the law of propositional calculus: ¬p implies q this is equivelant to p or q

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#7 h Apr 13, 2025, 4:51 PM

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"If you judge a fish because it cannot climb a tree , it will be foolish"-Unknown.

I am not commenting further in this post thank you.

Thanks to aops for moving it to MSM and I support it.

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SpeedCuber7

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#8 h Apr 13, 2025, 5:02 PM

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@triklino dude that's an awesome username i didn't even know greek letters were allowed lol

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#9 h Apr 13, 2025, 7:10 PM

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Proof: $x^2-5x=0$ Which means that the roots of this equation have to be real so we can use an method that is lost in the darkness called factoring. We can factor out the $x$ from each of the terms on the left hand side and get $x(x-5)=0$ which with more logic we can find that the possible outcomes is that if the Parantheses are 0 or the x=0 so first we can subsitute a value of x into that to make the value 0 so we get that x=5 and we finally get the solutions of $x=5$ and $x=0$ and to wrap up our proof we can prove that factoring is the best way to go because with quadratics you would only find 2 possibiliteis or 1 depending on the plus minus. And the other thing is that if you divide by x and just solve it with algebra then you will only get the solution of 5. Thus, proving that factoring is the best method out of all of them. We can use the remainder theorem to prove this which can be done easily. $\blacksquare$

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τρικλινο

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#10 h Apr 14, 2025, 2:56 AM

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please read the previous posts

The question here is not which is the best method to solve the problem,but if we lose a solution if we solve the problem by dividing the equation by a non zero x

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#13 h Monday at 1:22 PM

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Dear Triklino , I am high yesterday I am sorry for being rude. Can you please explain your Question properly that is what the thing you exactly want to know. According to what I understand, you wanted to know why in 2nd Solution $x=0$ is lost? right. Well forget about field theory and all that, Let me explain it in layman's term what is actually happening. In second solution , the solution $x=0$ is not actually lost. The reason we are getting $x=5$ but not $x=0$ is beacuse when we are dividing by $x$ we making an assumption that $x\neq 0$ and since we are making this assumption the solution $x=0$ is lost. For example when we divide by $x-5$ the solution $x=5$ is lost why $x-5=y(say)$ and we are assuming $y\neq 0$ which is equivalent to $x\neq 5$ . Now divison by zero is not possible which is not at all very easy to explain. Now $x=5$ and $x=0$ is not possible at the same time. Thus either $x=0$ or $x=5$ .

Now why I am talking about field beacuse $0$ and $5$ can be same when we are in a field of $\text{char} 5$ . If you are avoid knowing what is field that's perfectly fine to learn later, but just in layman's term note that $0=5$ is possible in finite fields of charecteristic $5$ .

Well I like sour grapes but fox will be happy if he clear your doubt thanks.

I hope it is clear now. If it is not then text me in dm.

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