For a quadratic with a letter in it, repeated roots means b² − 4ac = 0. Form the discriminant in terms of k, set it equal to zero and solve for k. For real roots, use b² − 4ac ≥ 0 instead.
This is the last lesson in quadratic structure and discriminants and ties together the discriminant, expansion and factorisation.
How is this different from the earlier discriminant lesson?
In using the discriminant for intersections you equated a line and a curve. Here the equation is given, but a coefficient is unknown. The discriminant becomes an expression in k, so you finish with another equation or inequality to solve.
Careful expansion matters more than usual. A single lost cross term will change every answer after it.
Steps
- Write the equation as ax² + bx + c = 0 and list a, b and c, each possibly containing k.
- Form b² − 4ac, bracketing every part that contains k.
- Expand and simplify into a polynomial in k.
- Set it to 0 for repeated roots, or use the inequality for other conditions.
- Solve and check each value by substituting it back into the original equation.
Worked example
Find the values of k for which x² + (k − 3)x + k = 0 has repeated roots. Then find the range of k for which it has no real roots.
Step 1: a = 1, b = k − 3, c = k.
Step 2, discriminant: (k − 3)² − 4(1)(k).
Step 3, expand: k² − 6k + 9 − 4k = k² − 10k + 9.
Step 4, factorise and set to 0: (k − 1)(k − 9) = 0, so k = 1 or k = 9.
Check k = 1: x² − 2x + 1 = (x − 1)², repeated root x = 1.
Check k = 9: x² + 6x + 9 = (x + 3)², repeated root x = −3.
No real roots: the discriminant is negative, so (k − 1)(k − 9) < 0. The product of these two brackets is negative between the roots, so 1 < k < 9. Check with k = 5: 25 − 50 + 9 = −16, which is negative.
The mistake to watch for
The common slip is to square (k − 3) as k² − 9.
Mistaken working: (k − 3)² − 4k = k² − 9 − 4k = 0, giving k = 2 ± √13
The middle term −6k was dropped when squaring the bracket.
Test it: k = 2 + √13 is about 5.6, which is inside the range 1 < k < 9 where the roots do not exist. The correction is to write (k − 3)(k − 3) in full and expand it term by term.
Check yourself
1. Find the values of k for which 3x² + kx + 12 = 0 has equal roots.
Show answer
a = 3, b = k, c = 12. The discriminant is k² − 4(3)(12) = k² − 144. Set it to 0: k² = 144, so k = 12 or k = −12. Check k = 12: 3x² + 12x + 12 = 3(x + 2)².
2. For what values of k does x² − 4x + k = 0 have distinct real roots?
Show answer
The discriminant is 16 − 4k. For distinct real roots we need 16 − 4k > 0, so k < 4. Check at k = 4: x² − 4x + 4 = (x − 2)², repeated, so k = 4 is the boundary and excluded.
3. Find the values of k for which kx² − 4x + k = 0 has repeated roots.
Show answer
a = k, b = −4, c = k. The discriminant is 16 − 4k². Set it to 0: k² = 4, so k = 2 or k = −2. Neither makes the equation linear. Check k = 2: 2x² − 4x + 2 = 2(x − 1)².
Where this leads next
After this module, continue to polynomial factors and remainders, then try the mixed practice set. The quadratic structure explorer lets you slide k and watch the discriminant change sign, and the non-calculator working trainer is good for the expansion arithmetic.
If algebra in a letter is where your confidence drops, our teachers can rebuild that step by step in online one-to-one Additional Mathematics tuition.