system of quadratic equations

[Steven G]

[QUOTE="homeschool girl, post: 485266, member: 77925"

in [MATH]x=\frac{5\pm \sqrt{25-12y}}{6}[/MATH] if [MATH]y>2[/MATH] then it will not be real.
[/QUOTE]
Please remember that there are numbers other than 1, 2, 3, .... After all, when you go to the store most items are NOT $1 or $2 or $3,...
Fine you want 25-12y<0. Then 25<12y and 25/12 < y or y>25/12 = 2 1/12 NOT 2. There are many numbers between 2 and 2 1/12 that satisfies your inequality would you left out.

 

[homeschool girl]

[MATH]x = \frac{4 \pm\sqrt{-4^2 - 4c}}{2}[/MATH]
-4^2 - 4c

16 - 4c

if c>4 then we get nonreal roots,

if c=4 there is only one set of roots,

and if c<4 there are multiple real roots?

 

[Steven G]

You have one problem -4^2 is not 16. Think of it this way. Someone put a negative sign in front of the number 4^2. Since 4^2 = 16 and if you put a negative sign in from of it you should get -4^2 = -16 NOT 16.

However you should have written (-4)^2 which means (-4)*(-4) which does equal 16. That is (-4)^2 =16. The formula b^2 - 4ac says to square b, not part of b or most of b but all of b. That means if the number that b equals has a negative sign then you better be squaring that negative sign as well.
Your two errors cancelled one other out so LUCKILY in the end you have the correct answer.

 

[Dr.Peterson]

Good work.

You did leave out some parentheses that make your work technically wrong; it should have been [MATH]x = \frac{4 \pm\sqrt{(-4)^2 - 4c}}{2}[/MATH]. But you didn't do what you wrote ([MATH]-4^2 = -(4^2) = -16[/MATH], whereas [MATH](-4)^2 = 16[/MATH].) So your answers are correct.

Just to illustrate (this is not part of the problem), suppose that c = 3. Then we expect to find more than one solution. Solving [MATH]x^2 - 4x + 3 = 0[/MATH], we actually find that we can factor it: [MATH](x-1)(x-3) = 0[/MATH], so x = 1 or 3; then, using the second given equation, [MATH]y = 2(1)^2 - (1) - 3 = -2[/MATH] when x=1, and [MATH]y = 2(3)^2 - (3) - 3 = 12[/MATH] when x=3. So the two solutions would be (1,-2) and (3,12).

If you made c=5, on the other hand, you'd find that there were no solutions.

But again, what you said was the answer:

if c>4 then we get nonreal roots,

if c=4 there is only one set of roots,

and if c<4 there are multiple real roots

Well, I'll make one little correction: it isn't "one set of roots"; it's "one root", which is an ordered pair. And I prefer the word "solution" to "root" in this situation; that's the problem's wording.

 

[homeschool girl]

You have one problem -4^2 is not 16. Think of it this way. Someone put a negative sign in front of the number 4^2. Since 4^2 = 16 and if you put a negative sign in from of it you should get -4^2 = -16 NOT 16.

However you should have written (-4)^2 which means (-4)*(-4) which does equal 16. That is (-4)^2 =16. The formula b^2 - 4ac says to square b, not part of b or most of b but all of b. That means if the number that b equals has a negative sign then you better be squaring that negative sign as well.
Your two errors cancelled one other out so LUCKILY in the end you have the correct answer.

oh well, I leaned that -4^2 implies that were asking for the square of the negation of 4 not the negation of the square of 4, and to specify negation of the square of 4 you need to write -(4^2) not the other way around, but thanks for correcting my "lucky mistake"

 

[lookagain]

[QUOTE="homeschool girl, post: 485266, member: 77925"

in [MATH]x=\frac{5\pm \sqrt{25-12y}}{6}[/MATH] if [MATH]y>2[/MATH] then it will not be real.

\(\displaystyle 3x^2 \ - \ 5x \ - \ y \ = 0\)

The radicand is incorrect, because you are repeating the incorrect expression that homeschool girl wrote. Each time she put values into
the quadratic formula, she showed them being substituted incorrectly and henceforth worked incorrectly. Yes, there was the partial
canceling of errors, because she thought \(\displaystyle \ -5^2 \ \) was the same as 25. In this case, a = 3, b = -5, and c = -y. So, the radicand for this
expression would be 25 + 12y.

 

[Steven G]

oh well, I leaned that -4^2 implies that were asking for the square of the negation of 4 not the negation of the square of 4, and to specify negation of the square of 4 you need to write -(4^2) not the other way around, but thanks for correcting my "lucky mistake"

When I writes -4^2 the mean -(4^2).

Think about this. How much is 20 -4^2. I suspect that you will say 4 because 20-16 = 4. Not that -4^2 = -16.

 

[Dr.Peterson]

oh well, I learned that -4^2 implies that were asking for the square of the negation of 4 not the negation of the square of 4, and to specify negation of the square of 4 you need to write -(4^2) not the other way around, ...

I don't want to make a big argument out of this; it is not a major part of your question. But I am interested.

It's possible that in some places it is taught that [MATH]-4^2[/MATH] means [MATH]-(4^2)[/MATH], but that is not true in any curriculum I know of. Can you show a page from a textbook that teaches this? I like to be aware of things that are taught differently in other places.

On the other hand, many students have told me they thought they were taught that.

 

[Steven G]

It's possible that in some places it is taught that −4² means −(4²).

Isn't that true everywhere?

 

[Dr.Peterson]

Isn't that true everywhere?

As far as I am aware -- but there's a lot of world out there, and I know enough about education in some distant parts of the world that I don't assume they all get it right! Even some teachers in America may go their own way ...

So I'm mostly being polite by not just saying "You're lying", but I'm also sincerely asking for evidence.