Linear and Quadratic Equation

[puzzleking123]

I have a linear equation and a quadratic equation, how (analytically) do I find the points at which they cross?

[factor-fiction]

Please?

Suppose the equations are:
y= ax² +bx +c and
y= mx +d
The solution occurs when ax² +bx +c = mx +d
Rewrite this as a quadratic:
ax² + (b-m)x +c-d = 0
Then solve the quadratic to find the solution x

Let me know your equations if you need further help.

Will you be acknowledging any of my answers and outstanding requests for clarification regarding earlier threads?

[factor-fiction]

Of course, once you find the solutions for x you substitute these into one of your equations to find the corresponding y values; this then gives you the coordinates of the points where the line and curve meet

[vascop]

Be careful because if the quadratic above has complex roots then this means that the line does not intersect the original quadratic.

[DaisyNonna]

Throw them in the air and note where they intersect

[factor-fiction]

Lol, vascop- puzzleking123 is a complex character for sure but I'd be very surprised if he is working with complex numbers here

[ChuckFickens]

Now that's were I started to get lost at maths in school, Imaginary and complex numbers!

[vascop]

I was just thinking that if PK just chose a quadratic and line at random then he could easily find that they do not intersect.

[nightmare]

vascop,
if a quadratic has complex roots, it merely means it does not meet or cross the x-axis.
It does not necessarily mean it will not cross the line given by the linear equation.

[vascop]

Please read what I said carefully!! I was not referring to the original quadratic, but the second quadratic ax^2+(b-m)x +c-d=0 in factor's original post. If the line and the ORIGINAL quadratic do not intersect when drawn on graph paper then the second quadratic WILL have complex roots. Try it and see.

[factor-fiction]

Maybe PK123 will come on to thank us and clarify the issue around complex roots.

[vascop]

Example 1
Quadratic is parabola y=x^2 (a=1 ,b=0, c=0)
Line is y=1 (line parallel to x axis)
This line cuts the parabola at the 2 points x=-1, y=1 and x=1, y=1
Using Factor's method m=0, d=1
So solve x^2-1=0 so x=1 and -1 as to be expected.

Example 2
Quadratic is parabola y=x^2 (a=1 ,b=0, c=0)
Line is y=-1 (line parallel to x axis)
This does not cut the parabola
Using Factor's method m=0, d=-1
So solve x^2+1=0 so x=i and -i (complex values) as expected when line does not intersect parabola

[nightmare]

sorry, vascop.
my brain was temporarily disengaged there.

[factor-fiction]

We'll probably never know. There seems to be a technical issue on AB which prevents puzzleking123 from accessing his threads after I have posted on them.

[vascop]

We should probably make a pact to ignore his/her posts unless he/she agrees to at least acknowledge answers

[factor-fiction]

Maybe. I've gone through phases of not answering PK123 but I sometimes can't resist in case others want to know the answer.
PK123 sometimes does acknowledge answers - but not mine for some reason.

[factor-fiction]

Well it's the last time I help this poster

[puzzleking123]

thanks for your help i've failed this module at school and will be doing resits at summer camp so I will be able to give you an answer then