# Thread: Solving quadratic equation with negative coefficient

1. ## Solving quadratic equation with negative coefficient

Hi - this is a Mom here trying to understand this so I can explain it to my kids, so any complete and EXPLICIT help would be greatly appreciated! I understand how to use the box method, and get everything correct when the coefficient is positive. However, when it becomes a negative, my answers are wrong everytime. I know that you factor out the negative 1 first, then do the box method. The question is, how do you know which set to multiply it back into?

For example: -x^2-3x+28. Please solve using box method.

Other example -m^2-13m-40

I get the answers correct, but my signage is always wrong on one of the groupings. Any help would be great!

2. Hi Mom:

I'm not familiar with any "box" method for solving quadratic equations.

However, I found the following "box" reference for factoring quadratic polynomials.

Since factoring is a common first step toward solving, perhaps this is what you mean.

THIS PAGE contains an example of how to handle a leading coefficient of -1.

If the information there does not help, then I'll need to see some of your work, in order to possibly determine what you're trying to do.

Cheers ~ Mark

3. ## Re: Solving quadratic equation with negative coefficient

Hi Mark - I went to purplemath, but I'm still not getting the right answers. For example:

-x^2 - 3x + 28.
I first factor out the negative one, so I get -1(x^2 + 3x -28)
I then make my square
with x^2 in the top left, -4x in the top right, 7x in the bottom left, and -28 in the bottom right.
I then factor going left and up.
This will give me (x+7) (x-4), assuming that you take the the positive or negative sign from the number to the left or on top. The answer in the book, however, is (-x+4)(x+7). So, it looks like you apparently take part of my answer (x-4) and multiply it by -1. But how do you know which one to multiply it to? Do you actually have to work it out for the original equation to see where the negative/positives must be once you have the factoring? I thought there was a way to know how to do this with the box method without have to then multiply everything out to find out where to "multiply" the -1. I just want to make sure I'm not doing the box method wrong. When the coefficient is positive, I get the answer right everytime. Have I missed some step, or do I actually need to multiply my answer from the box out to get the right answer?

Thank you so much for your help!
Suzanne

4. ## Re: Solving quadratic equation with negative coefficient

Originally Posted by asissa
Hi Mark - I went to purplemath, but I'm still not getting the right answers. For example:

-x^2 - 3x + 28.
I first factor out the negative one, so I get -1(x^2 + 3x -28)
I then make my square
with x^2 in the top left, -4x in the top right, 7x in the bottom left, and -28 in the bottom right.
I then factor going left and up.
This will give me (x+7) (x-4), assuming that you take the the positive or negative sign from the number to the left or on top. The answer in the book, however, is (-x+4)(x+7). So, it looks like you apparently take part of my answer (x-4) and multiply it by -1. But how do you know which one to multiply it to? Do you actually have to work it out for the original equation to see where the negative/positives must be once you have the factoring? I thought there was a way to know how to do this with the box method without have to then multiply everything out to find out where to "multiply" the -1. I just want to make sure I'm not doing the box method wrong. When the coefficient is positive, I get the answer right everytime. Have I missed some step, or do I actually need to multiply my answer from the box out to get the right answer?

Thank you so much for your help!
Suzanne
Your subject line says "solving quadratic equation....." but your examples are NOT equations. They are expressions. An equation MUST have an equals sign.

-x^2 - 3x + 28 = 0 IS an equation, and if you've got an equation, you can multiply both sides by the same non-negative number. So, if the negative coefficient on x^2 bothers you, you could multiply both sides of the equation by -1:

-1(-x^2 - 3x + 28) = -1(0)
x^2 + 3x - 28 = 0

And proceed from there.

I STRONGLY suspect that your problems involve factoring expressions.

Your first step of factoring out -1 is a good one. And you've correctly applied the "box method" to get

-1(x - 4)(x + 7)

Right there, you have your answer in factored form, and I would be tempted to leave it that way. OR, you could multiply (-1)*(x - 4) to get the "book's answer": (-x + 4)(x + 7). OR, you could multiply (-1)*(x + 7) to get yet a third answer: (x - 4)(-x - 7). I'm not sure that any one of those is "more correct" than the others.

5. ## Re: Solving quadratic equation with negative coefficient

Thanks! The book did not go over the box method (the teacher discussed it briefly in their class) and just gave the example by showing all possible ways it could be factored and then doing the math to see which one is correct. I find that a little inefficient and am trying to get my kids to use the box method which is quicker and makes a lot more sense. I thought maybe there was a "rule" that I was misunderstanding, because as you said, it doesn't matter which set I multiply by negative 1, the answer will be the same, but the book was only listing one correct answer, which baffled me. I appreciate you clearing it up for me. I sent them in to their teacher and told them to ask her specifically about the box method and the negative coefficient factoring this morning just to make sure. Thank you -

Suzanne

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