# Squaring a negative integer, math is wrong?

Discussion in 'General Discussion Forum' started by Ohaimerk, Aug 22, 2009.

1. ### OhaimerkCarbon Dated and ProudOrderite

Mar 30, 2009
[irrelevant backstory]
Started college yesterday, first class is college algebra, first assignment is solving quadratic equations by completing the square. In the quadratic equation (x^2)-6x+5=0, it is required to square a negative three when solving to find the answer.
[/irrelevant backstory]

The calculator told me that -3^2 = -9. But when you square a number, isn't that the same as (-3)*(-3)? I assume it would equal positive 9 because it's a negative times a negative. Research (can't find the link anymore) says that when you square a number you only apply the square to the integer because the negative sign becomes an implied 0-x, meaning that the way to square a number is 0-(3^2).

What am I missing? Is everything I know wrong? Is squaring a number not the same as taking it to the power of two?

2. ### Deleted member 53669Guest

It depends on what your type of mathematics you are doing. You probably have the calculator setting for calculus equations, where to solve the rate in variable change requires the use of multiplying negatives.

Say in normal mathematics and algebra (-2x-2)=4, which means that -2 has been multiplied by a negative, which will always change the value of a number towards it's opposite value...being a positive.

In calculus, -(2x-2)=-4, since you are multiplying a debt of numbers... as in increasing the amount of loss.

3. ### HamenaglarIt Wandered In From the Wastes

Jun 2, 2009
Here's an answer to your question. -3^2 = -9 is true. Why because it is: -(3*3) = -9.

However... (-3)^2 = 9.

When squaring the number it's always positive.

4. ### OhaimerkCarbon Dated and ProudOrderite

Mar 30, 2009
The text book is very poorly written and has only two simple examples of solving quadratic equations by completing the square, not even explaining what to do in a situation with complex numbers(which seems to be the case in at least half the problems so far) but I experimented with problems based on answers in the back and with calculators on the net, and it looks like my book(this section at least) treats a squared negative integer as -(3 * 3). Thanks for the replies duders, it's been five years since I set foot in a math classroom and I'm quite rusty.

5. ### HamenaglarIt Wandered In From the Wastes

Jun 2, 2009
Are you talking about a text book of your calculator or college text book?

Because if it's the latter, I'm sorry, but NO educational book in the world treats squared negative integer as -(3*3). It's just plain wrong. However, -3^2 is not squaring -3, it's squaring 3 and then adding minus infront. If you want to square -3 then it's meant to be written like (-3)^2.

P.s. The solutions to your quadratic equation should be 5 and 1.

6. ### OhaimerkCarbon Dated and ProudOrderite

Mar 30, 2009
It was the college text book, and it didn't. It showed how to do the problem, but it simplified without showing how it simplified and left many steps out, hence my intense confusion on what the hell was going on and my guessing and head exploding. When figuring the problem the calculator kept telling me that -3^2 was -9, and when I used that number the answer I came to never matched the answer in the back of the book. After doing the same problem for two hours trying to guess where I went wrong, I realized the whole problem with negative numbers being squared on my calculator, did some net research, and came here.

Yer
Code:
`(ax^2)+bx+c=0, where a=1, b=-6, and c=5`
Code:
`(x^2)-6x+5=0`
From the start, I subtract five from each side
Code:
`(x^2)-6x=-5`
This is where I got stuck. In this step, following the book's instructions, I divide the -6 by two and square the quotient, then add that number to each side(I don't know WHY exactly, it just works like that I guess). -6/2 is -3, and the resulting square, depending on which way I go, is either 9 or -9(which I now know the right answer to be 9)
Code:
`(x^2)-6x+9=-5+9`
Simplify:
Code:
`(x^2)-6x+9=4`
Factor it(I haven't factored in ages and I'm not sure I'm doing it right because the book just says to factor it, not going into how) But iirc I find factors of 9 that add up to -6, which in this case is (-3 * -3) because ((-3 + -3) = -6)
Anyways, completing the factoring:
Code:
`(x-3)(x-3)=4`
Simplify:
Code:
`(x-3)^2=4`
Take the square root of each side
Code:
`(x-3) = 2`
Move the three over and the answer is three plus-or-minus two(aka, 5, 1)

Anyone know of a site that explains WHY all those funky methods of number jumbling are applied, and/or a history of quadratic equations? Maybe if I can dive into the lives and minds of those who thought it up I can understand it better, because if I only know the HOW I'm not really learning anything that can't be done by a computer(and things like this interests me)

7. ### JayGreyIt Wandered In From the Wastes

Jun 8, 2009
"I'm not really learning anything that can't be done by a computer"

September 8th, I'll be back doing this stuff, after 3 years out of school. I wouldn't mind knowing the asnwer to this, either.

But think, if an EMP knocks out the computers of the world, you'll know. You'll know.

8. ### GuestGuest

You do know there's a simple formula for solving quadratic equasions? Or you guys supposed to do it the complicated way?

9. ### AshmoHalf-way Through My Half-lifeOrderite

Jul 2, 2004
To drive the point home: a possible error source would be if the calculator evaluated the expression "-3^2" as "-(3^2)" rather than "(-3)^2".

Other than that, yes, (-3)^2 is usually = 9 because sqrt(9) = [3;-3]

10. ### HamenaglarIt Wandered In From the Wastes

Jun 2, 2009
Oh my god, what a silly way to solve quadratic eqations. really silly.

Take a look at wikipedia.

http://en.wikipedia.org/wiki/Quadratic_equation (usually from my experience, wikipidia is a good source for these kinds of things).

if the eqation is a*x^2 + b*x + c. In your case a = 1, b = -6 and c = 5, you can apply the following formula.

x1 = (-b+sqrt(b^2-4*a*c))/(2*a)
x2 = (-b-sqrt(b^2-4*a*c))/(2*a)

where sqrt stands for square root.

P.s. what are you studying.

11. ### OhaimerkCarbon Dated and ProudOrderite

Mar 30, 2009
Yeah, you're using the quadratic formula. The other two ways are completing the square, and factoring. The way I am doing it is the way it will be on the test, which is completing the square

This class is college algebra, which is a requirement for a CIS degree(which is sorta like computer science for dummies, I slacked off too much in high school for anything better)

I have no idea how the calculator does it, I just know that when I enter -3 and press the square button and equals, it says it's -9, and also does the same when I use the exponent button. I'm still trying to figure it out(it's a ti-30xIIs, btw) but it looks like it has three settings for output, sci, eng, and flo, looks like I'm in flo mode. It also has settings for deg, rad, and grd(it looks like it's in deg mode)

Which settings should I use for this kind of math?

12. ### HamenaglarIt Wandered In From the Wastes

Jun 2, 2009
Who's your math teacher. Please introduce us, so that I can kill him. I always detest when you are expected to solve something exactly that precise way especially if there are simpler and faster options. I'm a computor science bachelor and I've never in my life had a need to solve equations the way you are being thought.

I'd recommend switching to sci, I assume that is scientific mode. Also if the problem is not solved by this, just put the brackets and input (-3)^2 in calculator instead -3^2. I'm not sure how your calculator works, but my bet is that it calculates -3^2 = -(3*3). Which isn't wrong, just not what you expect. Use brackets and it should be alright.

13. ### OhaimerkCarbon Dated and ProudOrderite

Mar 30, 2009
That's the magic of a community college education.

the sci setting looks like it just turns on scientific notation. But when I put the -3 in brackets and square it it comes out as positive 9. Thanks, I'll do that from now on