Question

Sorry if this doesn't make much sense, I'm not familiar with the english mathematical terms.

I have a quadratic formula that goes

2x^2+3x+c=0

And I also know that one of the solutions is 1.

Now I need to figure out how to determine what 'c' is in this equation.

If it helps any, I know that the answer is that 'c' is -5, but would of course like to know how to figure that out.

Answer 1

you know that 1 is a solution; therefore (x-1) has to be a factor

Answer 2

(x-1)(kx+n) = 2x^2+3x+c would be a good way to compare

Answer 3

let me edit that alittle ... dont need thos defulters :)

(x-1)(kx+n) = 2x^2+3x+c

kx^2 -kx +nx -n = 2x^2+3x+c ; k = 2 by default


2x^2 -2x +nx -n = 2x^2+3x+c

2x^2 +(n-2)x -n = 2x^2+3x+c

n-2 = 3

Answer 4

I'm of course not going to question whether that's right or not, but I'm not entirely sure how that answers the question? As stated before I know that the answer is that 'c = -5', and I don't really understand how you get that from your answer.

Answer 5

the only way 2 polynomials are equal is if they have the same parts ....

Answer 6

if 1 is a solution to the setup; then we know that (x-1) is a factor of the setp

(x-1), times something; is equal to 2x^2+3x+c

let that unknown something be a general linear factor .. (kx+n)

(x-1)(kx+n) has to have the same parts as 2x^2+3x+c

its just a matter of expanding it out and sorting thru the parts

Answer 7

(x-1)(kx+n) = x(kx+n) -1(kx+n)
= kx^2 +nx -kx -n
= kx^2 + (n-k)x -n
=2x^2 + 3 x +c

comparing parts we have:

k=2

n-k = 3

c = -n

Answer 8

Ahhh, I understand now. Thank you!

Answer 9

youre welcome :)