How to find range of values of p

[pancakes_fn]

Find the range of values of p for which the graph of
y = p(x+1) (x-3) - x + 4p + 2 does not intersect the x-axis.

I got to here:
y = px^2 - 2px + p - x + 2
How am I supposed to find the discriminant?

 

[pancakes_fn]

Does c refer to everything after the ax^2 + bx?

 

[Romsek]

\(\displaystyle y = px^2 - 2px + p - x + 2\\

y = px^2 -(2p+1)x + (p+2) \\~\\
a = p\\
b=-(2p+1)\\
c= p+2\\~\\
D = b^2 - 4ac
\)

 

[MarkFL]

Yes, write your quadratic as:

[MATH]px^2+(-2p-1)x+(p+2)=0[/MATH]
Now, require the discriminant to be negative...

 

[Steven G]

Does c refer to everything after the ax^2 + bx?

No, not at all! c is the constant term. For example in 3x^2+5x -9 +3x -11x^2, c is NOT -9 +3x -11x^2. In 5x-7+9x^2, c = -7

 

[pancakes_fn]

oh ok thanks all for help. im currently out and will get back soon

 

[pancakes_fn]

so p>1/4, right. thanks all

 

[MarkFL]

[MATH](2p+1)^2-4p(p+2)<0[/MATH]
[MATH]4p^2+4p+1-4p^2-8p<0[/MATH]
[MATH]-4p+1<0[/MATH]
[MATH]\frac{1}{4}<p\quad\checkmark[/MATH]