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

• Mar 8th 2011, 12:30 PM
iragequit
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?
• Mar 8th 2011, 12:37 PM
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 $\displaystyle \displaystlye x= \frac{-(b-m)^2\pm\sqrt{(b-m)^2-4a(c-k)}}{2a}$
• Mar 8th 2011, 01:03 PM
iragequit
Quote:

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
• Mar 8th 2011, 01:11 PM
topsquark
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
• Mar 8th 2011, 01:19 PM
iragequit
nbm