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Dividing by variable when solving algebraic equation
#1
Dividing by variable when solving algebraic equation
I know for all you math masters out there, this is basic stuff. But for people like me, it used to baffle me a little how dividing by a variable could sometimes lead to an incomplete solution set when solving an equation.

Consider the following for example:

x(x+1) = 0

The recommended approach to solving this is to use the zero factor property, leading to the solution set {-1, 0} for x.

But suppose, instead, you divided both sides by x, cancelling out the standalone x's on the left side of the equation. Then you're left with just:

x+1 = 0 which can only lead to the unique solution of -1 for x.
0 is no longer a solution.

I actually searched online for a website that could clearly explained why this happened, but couldn't find one. Maybe I used the wrong search keywords or something.

Anyway, I figured it out yesterday why this happens.

When you divide both sides of the original equation by x, the equation actually changes (meaning: it becomes a different equation with a different solution set for x).

Before dividing the equation by x, 0 was clearly one of the solutions.

But when you divide by x, 0 can no longer be a solution because then that would mean you were dividing by 0 in the case of x = 0! Division by 0 cannot work so x = 0 stops being a solution, and only -1 becomes the acceptable solution.

So rule of thumb when it comes to solving algebraic equations:

Don't divide by the variable when you have the option of using the zero factor property. Or you may end up with an incomplete answer.

I feel so happy finally getting this after all this time.
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#2
RE: Dividing by variable when solving algebraic equation
To be more exact, the reason the answer changes is because you're actually altering the original equation. The equation x(x+1) = 0 is a quadratic equation i.e. x^2 + x = 0. This is the form of a quadratic equation, which usually looks like ax^2 + bx + c = 0. Quadratics will always have a pair of solutions (and you can clearly see this when they are graphed - they intersect the x axis twice). But upon dividing by x, you are now left with a linear equation, and linear equations will only intersect the x axis once, hence only one solution.

EDIT: sorry, it's not entirely true that quadratics will always have two solutions. They can have one, and this happens when the curve is tangent to the x axis, meaning it's touching it exactly once.
"It is the mark of an educated mind to be able to entertain a thought without accepting it" ~ Aristotle
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#3
RE: Dividing by variable when solving algebraic equation
Nice Smile

Yes, you can't legitimately divide by a variable when the variable may be zero. In such a case, you've invalidated the whole thing. So you can check for whether the variable can be zero, and then treat the case where it's not zero, so you can divide by it.

You're not getting the zero solution this way because of this. Division by zero error. 0/0 doesn't equal 1. It is undefined.
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#4
RE: Dividing by variable when solving algebraic equation
Yeah, you need to assume that x=/=0 to divide by x.

And while we're at it, always remember that x^2 always has two solutions. Otherwise math teachers everywhere get mini heart attacks.


But my tears would drown the world, as my inner fire would reduce it to ashes.
Emil Cioran, On The Heights Of Despair
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#5
RE: Dividing by variable when solving algebraic equation
(25th October 2016, 07:48)Vic Wrote: Yeah, you need to assume that x=/=0 to divide by x.

And while we're at it, always remember that x^2 always has two solutions. Otherwise math teachers everywhere get mini heart attacks.

True (aside from 0, that is). On a related note, this includes non-real numbers as well. So another tip: don't forget the negative square roots of negative numbers when solving something like x^2 = -1.
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#6
RE: Dividing by variable when solving algebraic equation
It may have a repeated solution, however. Some quadratic equations have no real solutions, but they all have at least one complex solution.
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#7
RE: Dividing by variable when solving algebraic equation
(25th October 2016, 08:13)robvalue Wrote: It may have a repeated solution, however. Some quadratic equations have no real solutions, but they all have at least one complex solution.

Yes. The parabolae sometimes doesn't get to zero. Usually its when the delta under the square root is < 0
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#8
RE: Dividing by variable when solving algebraic equation
(25th October 2016, 07:48)Vic Wrote: And while we're at it, always remember that x^2 always has two solutions. Otherwise math teachers everywhere get mini heart attacks.

Man I don't think I ever really paid attention in any math class I ever took but I still remember that
"When life begins, we are tender and weak.
When life ends, we are stiff and rigid.
All things, including the grass and the trees,
Are soft and pliable in life, dry and brittle in death.
So the soft and supple are the companions of life,
Whilst the stiff and unyielding are the companions of death.
An army that cannot yield will be defeated.
A tree that cannot bend will crack in the wind.
Thus, by nature's own decree,
The hard and strong are defeated,
Whilst the soft and gentle are triumphant."
                                                           
                                                           -Lao-tzu


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#9
RE: Dividing by variable when solving algebraic equation
(25th October 2016, 07:48)Vic Wrote: Yeah, you need to assume that x=/=0 to divide by x.

And while we're at it, always remember that x^2 always has two solutions. Otherwise math teachers everywhere get mini heart attacks.

Now that I'm officially a math teacher, I can confirm that this is true. I got 3 bypass operations and emergency CPR twice just this afternoon.
The fool hath said in his heart, There is a God. They are corrupt, they have done abominable works, there is none that doeth good.
Psalm 14, KJV revised edition

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#10
RE: Dividing by variable when solving algebraic equation
(25th October 2016, 08:13)robvalue Wrote: It may have a repeated solution, however. Some quadratic equations have no real solutions, but they all have at least one complex solution.

Could I have an example?
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