Mathematics Problem - Algebra.

[BeautifullyDecayed.]

I am in level 2 [year 12] in New Zealand, you guys might not study this is USA or UK.. or wherever else but I'm hoping you might be able to help.
I am studying for a maths exam I have tomorrow.

I am stuck on questions about the discriminant, i know that this is b^2-4ac but have trouble applying it.
Here is an example of a question..

The roots of the equation x^2 + 6x + c = 0 are k and k − 1 .
Find the value of c.

I know that if the discriminant > 0 the equation has 2 real roots
if it is = 0 is has one real root
and if it is < 0 it has no real roots
also, if it is a perfect square the equation has two rational roots [what are these??]

So in this question it has two roots [k and k-1] so the discriminant must >0
in this case the discriminant would be
6^2 - 4 x 1 x c > 0

for this to be true, does c have to be > 9?

I get stuck from here..

 

[pingvincina]

Have you learned about Viete's formulas?

 

[BeautifullyDecayed.]

Not that I know of..
This is the formula sheet we are given...
http://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2008/90284-frm-08.pdf

 

[pingvincina]

If c>9 then D(discriminant)<0
If c<9 then D>0

 

[BeautifullyDecayed.]

Oh, right, thanks..
So would that be the answer?
c<9?
Or is it possible to put an exact number to it?

 

[pingvincina]

So the solution of this problem should be something like this:

x^2+6x+c=0
x1=k x2=k-1
That means this: x1=x2+1
(To shorten the solution I wil use D instead of 36-4c)

x1= (-6 + D)/2
x2= (-6 - D)/2

From x1=x2+1 we get:
(-6 + D)/2= (-6 - D)/2 +1

You should know what's next now...let me know the result

 

[pingvincina]

I was writing the post while you posted yours. Yes, as you can see from my post, it's possible to put an exact number to c

 

[BeautifullyDecayed.]

This may sound very stupid
but how did you get this..
x1= (-6 + D)/2
x2= (-6 - D)/2

I understand everything up to this point.

 

[pingvincina]

x1 = -b + sqrt(b^2 -4ac) /2a
x2 = -b - sqrt(b^2 -4ac) /2a
And then you just throw in the numbers you know.
a=1
b=6
And you get:
x1= -6 + sqrt(36-4c) /2
x1= -6 - sqrt(36-4c) /2

 

[BeautifullyDecayed.]

Great, thanks
I got c=8.75

 

[nitrocan]

k + k -1 = -6
k = -5/2
c = k*(k-1) = -5/2*(-7/2) = 35/4

 

[pingvincina]

c=k*(k-1)=35/4
She already wrote that she doesn't know about Viete -.-''

 

[BeautifullyDecayed.]

does c=8.75?

 

[nitrocan]

Then let me explain:
a(x-x1)(x-x2) = 0
a(x^2 -x*x2 - x*x1 + x1*x2) = 0
a(x^2 -x*(x1+x2) + x1*x2) = 0
so b = a*-(x1+x2)
x1+x2 = -b/a
and so x1*x2 = c/a

 

[pingvincina]

@Nitrocan: If they're still learning about determinant that means that they are on the beginning of quadratic equation. That means that she probably didn't learn about that decomposition.
@BeautifullyDecayed.: Yes that is the correct answer

 

[nitrocan]

@Nitrocan: If they're still learning about determinant that means that they are on the beginning of quadratic equation. That means that she probably didn't learn about that decomposition.
@BeautifullyDecayed.: Yes that is the correct answer

?

The formula b^2 - 4ac requires more math knowledge than the Viete thing or whatever.,
Besides, what is that formula sheet given with you? Are you supposed to look at the formulas there and solve the equations with them? Can you not use any other methods?

For example:
x^2 + 6x + c = 0
x^2 + 6x + 9 = 9 - c
(x+3)^2 = 9-c
x+3 = sqrt(9-c)
x = sqrt(9-c) - 3 and
x = -sqrt(9-c) - 3
x1 - x2 = 1 so
sqrt(9-c) - 3 + sqrt(9-c) + 3 = 1
sqrt(9-c) = 1/2
9-c = 1/4
c = 35/4

 

[pingvincina]

Sorry I had other things on my mid so i wrote determinant...Of course I meant discriminant*. I'm not saying that they are not allowed to use other methods. It's just that they are learning something new and they have to work it out by solving problems. That's all. And when they have better understanding and knowledge of quadratic equation they can move on freestyle i.e. using their knowledge to solve it in the shortest and most elegant way(like the one you wrote, which I like very much)

 

[nitrocan]

Sorry I had other things on my mid so i wrote determinant...Of course I meant discriminant*. I'm not saying that they are not allowed to use other methods. It's just that they are learning something new and they have to work it out by solving problems. That's all. And when they have better understanding and knowledge of quadratic equation they can move on freestyle i.e. using their knowledge to solve it in the shortest and most elegant way(like the one you wrote, which I like very much)

I didn't mean to sound like I was saying the "look at the formulas" stuff to you. I just don't like the way math is being taught. Showing how the formulas are derrived will make it so that the student will never have a problem with solving those kinds of equations again. It's like using no intuition on F2L, only algorithms (if that makes any sense ).

 

[pingvincina]

Hehe...That's a good comparison...I agree with your opinion about teachers teaching formulas to kids because by solving so many problems(''practicing'') they lose creativity...But it's enough spamming now I think

 

[Dene]

nitrocan: you overestimate the New Zealand education system - we don't learn to do maths in that way (although if you want to do scholarship maths in the final year it is necessary which is kind of stupid as it isn't taught).

BeautifullyDecayed: Don't worry too much; 6th form maths is a joke. Just make sure you put down anything you know, even if you only put down the formula that you are meant to use without applying it in any way. You get bonus points for that (who knows why, but take advantage of the moronic system as much as you can). Good luck for your exam tomorrow.