A parameter is a quantity whose value controls the behaviour of a whole family of equations or graphs. Quadratic parameter problems ask you to determine which values produce a required structure: two roots, one repeated root, no real roots, tangency or a specified intersection pattern.
The key is to translate the verbal condition into mathematics before solving for the parameter.
1. Build the condition dictionary
- Two distinct real roots → discriminant > 0.
- One repeated real root → discriminant = 0.
- No real roots → discriminant < 0.
- A line tangent to a quadratic → the intersection equation has a repeated root.
- Two graph intersections → the intersection equation has two distinct real roots.
2. Worked example: repeated roots
Find k if x²+(k−2)x+9=0 has equal roots.
Set Δ=0: (k−2)²−36=0. Thus k−2=±6, giving k=8 or k=−4.
Both values should be retained unless another condition removes one.
3. Worked example: two distinct roots
Find the range of k for x²−2kx+k+3=0 to have two distinct real roots.
Δ=(−2k)²−4(k+3)=4k²−4k−12=4(k²−k−3). Require k²−k−3>0.
The boundary values are k=[1±√13]/2. Since the quadratic in k opens upward, the required range is k<(1−√13)/2 or k>(1+√13)/2.
4. Tangency between a line and parabola
Find k if y=x²+kx+1 is tangent to y=3x−2.
Equate outputs: x²+(k−3)x+3=0. Tangency requires Δ=0, so (k−3)²−12=0. Therefore k=3±2√3.
5. Find the point of tangency as well
When the intersection quadratic has a repeated root, its x-coordinate is −b/(2a). For the previous example, x=−(k−3)/2.
If k=3+2√3, then x=−√3. Substitute into y=3x−2 to obtain y=−3√3−2. The other parameter value gives the other member of the quadratic family and a different tangency point.
6. Parameters can control whether roots are positive
The discriminant alone tells you whether roots are real, not whether they are positive. For a quadratic ax²+bx+c with real roots r and s, use r+s=−b/a and rs=c/a alongside the discriminant when sign conditions matter.
For both roots to be positive, a common route is: real roots, positive sum and positive product. The exact conditions should be checked against the particular quadratic, including repeated-root wording if relevant.
7. Parameter conditions from a turning point
For y=x²−4x+k, completing the square gives y=(x−2)²+(k−4). The minimum value is k−4.
If the graph must remain above the horizontal axis for all real x, require k−4>0, so k>4. If touching the axis is allowed, use k≥4.
8. The same condition can have two routes
For x²−4x+k>0 for every real x, discriminant reasoning gives 16−4k<0, hence k>4. Completed-square reasoning gives the same condition from the minimum.
When two independent representations agree, confidence in the result increases.
9. Common mistakes
- Solving for x when the question asks for k.
- Using Δ≥0 when “two distinct roots” requires Δ>0.
- Using Δ=0 before equating two graphs in a tangency problem.
- Assuming real roots are automatically positive.
- Discarding one parameter solution without a stated condition.
10. Practice
- Find k if x²+kx+16=0 has equal roots.
- Find the range of k for x²+2x+k=0 to have no real roots.
- Find k if y=x²+k is tangent to y=6x−5.
- Find the range of k for y=x²−2x+k to remain strictly above the horizontal axis.
- For x²−(k+1)x+k=0, verify that x=1 is always a root and find the other root.
11. Answers
- k²−64=0, so k=±8.
- Δ=4−4k<0, so k>1.
- x²−6x+(k+5)=0; Δ=36−4(k+5)=0, so k=4.
- y=(x−1)²+(k−1), so k>1.
- Substitution confirms x=1. Factorisation gives (x−1)(x−k)=0, so the other root is k; when k=1 the roots coincide.
Continue with Discriminants in Additional Mathematics, Quadratic Inequalities, or return to the Additional Mathematics Hub.