Simple Functions Question

Sep 2009
148
2
I've solved a function to this answer:

\(\displaystyle x^2-2x-kx+4=0\)

How would you convert the above into \(\displaystyle ax^2+bx+c\) format?

I've tried it like this

For bx: I've tried converting -2x-kx into \(\displaystyle -x(2+k)\) by taking -x common, but that's not giving me the right roots.

What am I doing wrong? What's the rule here to form bx?

Thanks!
 
Jun 2009
806
275
What are the conditions for the roots.
Since k is unknown, roots will be in terms of k.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
I've solved a function to this answer:

\(\displaystyle x^2-2x-kx+4=0\)

How would you convert the above into \(\displaystyle ax^2+bx+c\) format?

I've tried it like this

For bx: I've tried converting -2x-kx into \(\displaystyle -x(2+k)\) by taking -x common, but that's not giving me the right roots.

What am I doing wrong? What's the rule here to form bx?

Thanks!
What you have is certainly correct: \(\displaystyle x^2- 2x- kx+ 4= x^2- (2+k)x+ 4\).
 
Sep 2009
148
2
If this is correct then for the discriminant formula a, b and c would be:

a= 1
b= -(2+k) or -2-k
c= 4

Right?