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forewords: Honestly, I've tried looking around for any answer to this question, but it's so specific i can't find any. If it's already answered i apologise.

Alright so i have this equation $$-x^{2}+4=0$$

Now usually i would remove the negative in front of the polynomial, by multiplying the whole equation by $\frac 1{-1}$ or simply said, divide by $-1$. If we do this, we get.

$$x^2-4 = 0$$

Now if we factor the first equation we get. $$-x^{2}+4=-(x+2)(x-2)$$ If we factor the other we get $$x^2-4=(x+2)(x-2)$$ Ala the same, just without the minus. Graphing these two equations, yield different result (one is a frown, and the other is smiley). Could somebody please give me a intuitive explanation of why i should not divide by the minus, and instead put it outside the parenthesis Usually when i this far down the rabbit hole, the answer is humiliatingly simple, but now i can't seem to do anything right.

Any help is greatly appreciated, thank you.

Edit

I updated with a picture to show, where my confusion lies (how to divide by -1, and still get the same graph) Calculations

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Aron
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    It's okay that they give different graphs, as we only care about the roots, because they give the same roots we are good. – Ahmed S. Attaalla Oct 08 '16 at 17:00
  • I'd like them to give the same results, because i use a technique where i mark off the roots in a number line, and find out where they increase/decrease. Thereafter i am able to draw the graph, if a simple division by -1 messes up my equation the graph won't be the same, and I'm left with the wrong answer (but not the wrong roots). – Aron Oct 08 '16 at 17:05
  • If it is the case where you need to graph $y=x^2-4$. Then you can not multiply or divide or add or subtract almost anything to $x^2-4$ because that would change it., thereby changing it's graph Multiply,divide,add, subtract to both sides is method we use to solve equations, not graph functions. All we can do is multiply/divide by $1$, add or subtract by $0$, or factor, because these do not change $x^2-4$. @Aron – Ahmed S. Attaalla Oct 08 '16 at 17:10

2 Answers2

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It's okay to divide by $-1$.

It's true that the graph of $y=x^2-4$ is different from the graph of $y=-x^2+4$. But we're specifically interested in when the right-hand-side ($x^2-4$ or $-x^2+4$) is zero. And the two graphs cross the $x$-axis in exactly the same places, which means that $x^2-4$ and $-x^2+4$ are zero at exactly the same values of $x$.

Micah
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I suppose you are plotting $$y = -x^2 + 4 \quad\text{and}\quad y = x^2 - 4,$$ which represent parabolas, not $$-x^2 + 4 = 0 \quad\text{and}\quad x^2 - 4 = 0.$$ If you divide by $-1$ both sides of one of the former equations, you get $-y$ instead of $y$, from which you should see that they are not equivalent (i.e., they do not represent the same parabola).

Luca Bressan
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  • Oh god, thank you. I can't believe i didn't look at it that way. So suppose want to use my roots, and find out where the parabolas have a positive and negative value. Would i then have to just always keep in the back of my head that i divided it by -1, or is there some rule around this procedure. – Aron Oct 08 '16 at 17:07
  • You don't have to keep the $-1$ in the back of your head. If you want to solve $y > 0$, i.e. $-x^2 + 4 > 0$, you can multiply both sides of the inequality by $-1$ and change the direction of the inequality, i.e. write $x^2 - 4 < 0$. The only rule you need to remember is that whenever you multiply or divide by a negative number, you have to change the direction of the inequality. – Luca Bressan Oct 08 '16 at 17:13
  • Thank yo so much @Luacas, your answered really cleared things up for me. But i want to stretch it an inch further, before i let go of your helpfulness. Would you mind taking another look at the photo i added, it shows where my confusion lies, and the number line i keep talking about. Thank you greatly. – Aron Oct 08 '16 at 17:29
  • If the inequality is $-x^2 + 4 < 0$, the solutions of the inequality are given by the intervals represented by a dotted line, because you want to look at those $x$ for which $-x^2 + 4$ is negative, i.e. $(-\infty, -2) \cup (2, \infty)$. When you multiply by $-1$ and change the direction of the inequality, the solutions will be given by the intervals represented by a continuous line, because you want to look at those $x$ for which $x^2 - 4$ is positive, i.e. $(-\infty, 2) \cup (2, \infty)$. Thus the solution set is the same. – Luca Bressan Oct 08 '16 at 17:36
  • Thank you @Luca, i get it now. Your help was very useful. – Aron Oct 08 '16 at 17:51