# Thread: intersection of y=mx+k and y=ax^2+bx+c where a=/=0

1. ## intersection of y=mx+k and y=ax^2+bx+c where a=/=0

determine intersection of y=mx+k and y=ax^2+bx+c where a=/=0

i got

((-b+m+sqrt(-4a(c-k)+(b-m)^2))/2a, (m(-b+m+sqrt(-4a(c-k)+(b-m)^2)+k)/2a))

and

((-b+m-sqrt(-4a(c-k)+(b-m)^2))/2a, (m(-b+m-sqrt(-4a(c-k)+(b-m)^2)+k)/2a))

as my 2 solutions

is it correct?

2. Its hard to decodee your solution but it looks like your on the right track.

For $\displaystyle \displaystlye mx+k= ax^2+bx+c \implies 0=ax^2+(b-m)x+c-k$

Therefore $\displaystyle \displaystlye x= \frac{-(b-m)^2\pm\sqrt{(b-m)^2-4a(c-k)}}{2a}$

3. Originally Posted by pickslides
Its hard to decodee your solution but it looks like your on the right track.

For $\displaystyle \displaystlye mx+k= ax^2+bx+c \implies 0=ax^2+(b-m)x+c-k$

Therefore $\displaystyle \displaystlye x= \frac{-(b-m)^2\pm\sqrt{(b-m)^2-4a(c-k)}}{2a}$
why woudl you square -(b-m)? i'm gusesing that's a mistake right?

also, shoudl i leave it at -(b-m) or simplify / expand to -b+m +- sqrt(...etc..etc) ?

also, what's y equal to?
i subbed in x into y=mx+k and got y= m(-b+-sqrt(-4a(c-k)+(b-m)^2)/2a+k

i don't know how to simplify it after you sub it in

also it would be helpful if you could answer the rest of the questions in first post, thanks

4. Yes the $\displaystyle -(b - m)^2$ is a typo.

Take a look at the discriminant. Is this always non-negative? This is your extra criterion for the existence of a solution.

Just a thought. I wouldn't worry too much about an explicit form for y. Calculate and simplify x then feed that value of x into y = mx + k. Algebraically it will be simpler to deal with (as well as easier to write your solution.)

-Dan

5. nbm