# Equations with multiple solutions

1. Sep 20, 2015

### LogarithmLuke

Given the equation (x+4)(2-x)=x+4

How come it's wrong to divide both sides by x+4? Should it not give the same solution as multiplying out the left side and solving the quadratic equation?

2. Sep 20, 2015

### Mentallic

When you divide by a factor, you have to keep in mind that you can't divide by zero. If you do divide by zero which happens in this case when x+4=0, or x=-4, then you've made the assumption that x+4 does not equal zero, or $x\neq -4$ and hence you've lost the solution x=-4. This is why you learn quadratics and factoring, because each factor is a solution, and in the same way that you wouldn't ignore one of the factors, you also don't divide by any factors.

You could always expand the left hand side, rearrange everything into the standard quadratic form and either factorize again or use the quadratic formula, but much more simply, you can notice that

$$(x+4)(2-x)=x+4$$
$$(x+4)(2-x)-(x+4)=0$$
$$(x+4)(2-x-1)=0$$
$$(x+4)(1-x)=0$$

Notice from the 2nd to 3rd line, what I did there is essentially factoring out x+4 from both factors. It might seem more obvious if we make the substitution u=x+4 and then we have to factorize u by doing
$$u(2-x)-u$$
$$=u((2-x)-1)$$
$$=u(2-x-1)$$

Hence each factor equals 0 and so you have the two solutions x+4=0 which gives x=-4, and 1-x=0 which gives x=1.

Dividing through by x+4 would give us a linear equation which would only have the solution x=1 (because we've tossed out the x=-4 solution).

3. Sep 20, 2015

### LogarithmLuke

I didn't quite catch the factoring part. It looks to me as if you changed x+4 into -1. I assume what you did was basically the same as multiplying the two parantheses and subtracting by x+4, but i don't quite understand how you're able to do it like that. Are there more steps to the factoring, or am i missing something?

Also, when i divide both sides by x+4, i get that x=2. Shouldnt i atleast get one of the solutions? 2-x=0, x=2

Oh and btw, in step 2, why cant you do x+4 -(x+4) and get 0? This way you aren't dividing by 0.

4. Sep 20, 2015

### Dr. Manoj

I don't think the given equation satisfies LHS =RHS, when you substitute a value for given other than 1, LHS!=RHS

5. Sep 20, 2015

### Ssnow

If you want divide by $x+4$ you must assume that $x\not=-4$ because $-4$ is just a solution of the equation ...

6. Sep 20, 2015

### LogarithmLuke

How come you have to think about whether or not x+4=0? You're supposed to find out when left side is equal to x+4, not when it is equal to 0.

7. Sep 20, 2015

### Ssnow

If you have this equation $a\cdot b=a$ with $a,b$ numbers you can deduce by equivalence principles that $b=1$ if $a\not=0$. In fact if $a\not=0$ you can divide by $a$ both sides of the equation and obtain $b=1$.

8. Sep 20, 2015

### LogarithmLuke

I think i understand a bit more now, but a lot of this is still somewhat unclear to me. Namely the questions i listed in post number 3 of this thread.

9. Sep 20, 2015

### Staff: Mentor

In general, no. If you expand the left side and move all of the terms to the left side, you get a quadratic equation whose solutions are x = -4 or x = 1.

If you divide both sides by x + 4, you are tacitly assuming that x +4 is not zero, or in other words, that x ≠ -4. In this case, since x = -4 is a solution of the quadratic equation, you have "lost" one of your solutions.

You have made an error. If you divide the left side by x + 4, you get 2 - x. If you divide the right side by x + 4, you get 1. I suspect that when you divided the right side, you got 0.
I'm not sure what you're referring to here.

10. Sep 20, 2015

### LogarithmLuke

Yeah, i forgot that (4+x)/(4+x)=1/1

I was referring to Mentallics answer(post #2). (x+4)(2-x)-(x+4)=0 It seems to me as if you could cancel out the two parantheses with x+4. I also still don't understand the factoring done here. If you substitute x+4 for U,how come you change it into 1?

Oh and btw, thanks for all the help on this everyone:)

11. Sep 20, 2015

### Staff: Mentor

I don't know what you mean "cancel out the two parantheses with x + 4".

I'm not a fan of the technique of cancelling, as it obscures whatever operation is being performed.
BTW, the word is parentheses.
In this line, (x+4)(2−x)−(x+4)=0
Mentallic replaces x + 4 with u, to get u(2 - x) - u = 0
Then he brings out a factor of u from the last term, to get u(2 - x - 1) = 0, or u(1 - x) = 0
In my opinion, this substitution doesn't help to clarify things, and is more complicated than it needs to be.

12. Sep 20, 2015

### LogarithmLuke

I meant that you perform the action (x+4)-(x+4) and get 0. I'm sorry for any wrong terminology, English is not my first language.

I really struggle with understanding this factoring part. WIth or without substitution.

What is actually meant by "factoring out x+4 from both factors"? Is there perhaps a way to show this factoring in more steps, to make it more understandable? It seems like a very neat trick, and it would be nice to be able to apply this technique into future math problems.

13. Sep 20, 2015

### Staff: Mentor

And this is correct, but most likely the step leading to (x + 4) - (x + 4) is incorrect.
It's really nothing more than the distributive law, which says that a(b + c) = ab + ac. When you go from the left side to the right, you are expanding (or multiplying out) the factors. When you go from the right side to the left, you are pulling out the common factor (a) from both terms.

Some examples:
3*7 + 7.
Both terms (3*7 is one term and 7 is the other term -- terms are expressions that are added or subtracted) have a factor of 7.
So 3 * 7 + 7 = 7(3 + 1)
As a check, expand the right side to get 7 * 3 + 7 * 1 = 21 + 7 = 28

x2 + 3x -- common factor is x
x2 + 3x = x(x + 3)

(x + 4)(2 - x) + (x + 4) -- common factor is x + 4
(x + 4)(2 - x) + (x + 4) = (x + 4)(2 - x + 1) = (x + 4)(1 - x)
This is similar to, but not the same as, the substitution that Mentallic did

14. Sep 20, 2015

### LogarithmLuke

So, in step two you could do (x+4)-(x+4)? Wouldnt this eliminate one of the solutions again?

I'm familiar with factoring using parentheses, i just didn't understand that that's what he did here. So if i understand correctly, when you multiply x+4 with -1 you get -(x+4) just like you started with, which is why it was done.

Last edited by a moderator: Sep 20, 2015
15. Sep 20, 2015

### MrAnchovy

This seems to have got really complicated but it should not be.
• In the first equation you have (x+4)(something)=(x+4). It can clearly be seen from this that (something)=1 is a solution i.e. (2-x)=1, x=1.
• Now we all know that we cannot divide by zero, so what happens when x+4=0? We have 0(something)=0. This is always true, so x+4=0 i.e. x=(-4) is also a solution.

16. Sep 20, 2015

### Staff: Mentor

That doesn't make any sense. In the first line above, it is (x+4)(2-x)=x+4. You can't just subtract x + 4 from both sides to get 2 -x = 0. I think that's what you're trying to do.

17. Sep 20, 2015

### LogarithmLuke

I believe that dividing by zero is only a problem when it gives 0 in the denominator of the fraction, or is it the same here? I know that if you for instance have an equation with x in the denominator(s) you have to find the solutions which give 0 in the denominator(s) and mark these as false solutions.

Why is this? You have a positive apple and a negative apple, and together they equal 0. After you move x+4 over to the left side you don't have to subtract on both sides of the equation. Is it that (x+4)(2-x) has to be multiplied out first before you can subtract anything?

18. Sep 20, 2015

### Mentallic

Yes, BUT if you divide by x+4 and then check when x+4=0, giving us x=-4, you now have to assume it's not a solution, but it originally was a solution in the original equation, so now you've lost this x=-4 solution in your new equation (the equation changed once you divided through).

It's very similar to having the quadratic finally factored as
$$(x+4)(1-x)=0$$

and now at this point you decide to divide through by x+4, leaving us with 1-x=0 which only has the solution x=1. We've lost x=-4 and the whole point of factoring a quadratic and NOT dividing through by one of its factors is to ensure that we get both quadratic solutions. This is why we factor and never divide one of its factors. Don't confuse this with dividing by a constant number that isn't 0, because this is always ok to do.

No, the analogy would go more like this: You have a box of apples and subtract an apple, how many apples do you have left? Or, you have a slice of an apple and subtract an apple, how many left?

Yes, but like I said earlier, you don't need to multiply out in this case.

The distributive law is

$$ab+ac=a(b+c)$$

and so, letting c=-1
$$ab-a=a(b-1)$$
which is why
$$(x+4)(2-x)-(x+4)=(x+4)(2-x-1)$$

19. Sep 20, 2015

### Staff: Mentor

I believe that the mistake you are making is in thinking that (x + 4)(2 - x) - (x + 4) is equal to 2 - x. That is NOT true. Since the factor (2 - x) is MULTIPLIED by x + 4, the only way to get rid of the (x + 4) factor is to divide by it, not subtract it.

20. Sep 21, 2015

### LogarithmLuke

So it's kinda the same thing as factoring a quadratic function, then dividing by one of the factors on both sides of the equation. This would technically not be wrong to do, but would not be very smart as the whole point of factoring a quadratic is to find the x interecepts. I think i understand alot more now, thanks for all the help:)

21. Sep 21, 2015

### Dr. Manoj

The roots of equation are, (-4) \$ (1) it's better if you simplify the given equation. And solve using quadratic formula. In general, if you substitute value of x, surely you can divide both the sides. Got it?

22. Sep 21, 2015

### Staff: Mentor

For this problem, factoring is a perfectly valid way to solve the equation
I'm not sure what you're saying here. If you divide both sides of an equation by a factor that is associated with a solution, you are essentially dividing by zero.

The problem that the OP had difficulty understanding was why he couldn't divide both sides of (x + 4)(1 - x) = 0 by x + 4. Doing this resulted in 1 - x = 0, which is not equivalent to the equation he started with.