Equations » Quadratic equations with a parameter
This page is about quadratic equations with a parameter. You might want to read the general theory about quadratic equations first. There, you will also find theory about the quadratic formula (abc-formula) and the discriminant (D = b2 – 4ac).
In these exercises, you always have to calculate for which p the equation or formula of the parabola has only one point in common with the horizontal axis or another graph. When graphs have one point common they often touch instead of intersect.
When graph(s) have one point in common, the discriminant must be zero (D = 0).
If necessary, reduce the right-hand side of the equation to zero and make the equation for the discriminant.
Example 1
Calculate for which value(s) of p function f (x) = 2x2 + 5x + p touches the horizontal axis.
Answer:
2x2 + 5x + p = 0
Make the equation for D = 0.
| b2 – 4ac | = 0 |
| 52 – 4 · 2 · p | = 0 |
| 25 – 8p | = 0 |
| –8p | = –25 |
| p | = 318 |
Example 2
Calculate for which value(s) of p functions f (x) = 4.5x2 + px and g(x) = –2 touch.
Answer:
4.5x2 + px = –2
4.5x2 + px + 2 = 0
Make the equation for D = 0.
| b2 – 4ac | = 0 |
| p2 – 4 · 4.5 · 2 | = 0 |
| p2 – 36 | = 0 |
| p2 | = 36 |
| p = –6 | or p = 6 |
Example 3
Calculate for which value(s) of p functions f (x) = –x2 + 2x + 1.5 and g(x) = –4x + p touch.
Answer:
–x2 + 2x + 1.5 = –4x + p
–x2 + 6x + 1.5 – p = 0
Make the equation for D = 0. Note: c = (1.5 – p).
| b2 – 4ac | = 0 |
| 62 – 4 · –1 · (1.5 – p) | = 0 |
| 36 + 4(1.5 – p) | = 0 |
| 36 + 6 – 4p | = 0 |
| 42 – 4p | = 0 |
| 4p | = 42 |
| p | = 10.5 |
Example 4
Calculate for which value(s) of p the graph of function f (x) = –2x2 + px – 3 is entirely below the x-axis.
Answer:
You want NO intersections. Therefore, you must solve D < 0.
–2x2 + px – 3 = 0 (Actually < 0 because you want the graph below 0)
Make the equation for D < 0.
| b2 – 4ac | < 0 |
| p2 – 4 · –2 · –3 | < 0 |
| p2 – 24 | < 0 |
| p2 | < 24 |
– < | p < |
Check quadratic inequalities.
Example 5
Calculate for which value(s) of p the equation px2 + px – 2x + 4p = 0 has exactly one solution.
Answer:
First write the equation as ax2 + bx + c = 0.
px2 + px – 2x + 4p = 0
px2 + (p – 2)x + 4p = 0
Make the equation for D = 0.
| b2 – 4ac | = 0 |
| (p – 2)2 – 4 · p · 4p | = 0 |
| (p – 2)(p – 2) – 16p2 | = 0 |
| p2 – 4p + 4 –16p2 | = 0 |
| –15p2 –4p + 4 | = 0 |
| D = (–4)2 – 4 · –15 · 4 = 256 | |
p = –(–4) +  2 · –15 | of p = –(–4) – 2 · –15 |
| p = – 23 | of p = 25 |