The discriminant b² − 4ac tells you how many real roots a quadratic has. Positive gives two different roots, zero gives one repeated root, and negative gives none.
This lesson is part of quadratic functions. It follows relating roots to coefficients.
What do the three cases look like?
| b² − 4ac | Roots | Graph |
|---|---|---|
| Greater than 0 | Two different real roots | Crosses the x-axis twice |
| Equal to 0 | Two equal roots | Touches the x-axis once |
| Less than 0 | No real roots | Does not meet the x-axis |
The discriminant is the part under the square root in the quadratic formula. When it is negative, the square root is not a real number, which is why no real roots exist.
Worked example: a value of k for equal roots
Here is an original question. The equation x² + kx + 9 = 0 has equal roots. Find the values of k.
- Identify a = 1, b = k, c = 9.
- Equal roots means b² − 4ac = 0, so k² − 4(1)(9) = 0.
- Simplify: k² − 36 = 0, so k² = 36.
- Solve: k = 6 or k = −6.
Check with k = 6: x² + 6x + 9 = (x + 3)², which has a repeated root at −3. With k = −6: x² − 6x + 9 = (x − 3)², repeated root at 3. Both work.
A range of values
Find the range of m for which x² − 4x + m = 0 has real roots.
Real roots means b² − 4ac ≥ 0, so 16 − 4m ≥ 0. Then 16 ≥ 4m and m ≤ 4.
The boundary m = 4 gives the equal-roots case, (x − 2)² = 0. So it is included.
A tangent line
A line y = 2x + c is tangent to the curve y = x² + 3. Find c.
- Set the expressions equal: x² + 3 = 2x + c.
- Rearrange to zero: x² − 2x + (3 − c) = 0.
- A tangent touches once, so b² − 4ac = 0: (−2)² − 4(1)(3 − c) = 0.
- Simplify: 4 − 12 + 4c = 0, so 4c = 8 and c = 2.
Check: x² − 2x + 1 = 0 gives x = 1, so the touching point is (1, 4). The line y = 2x + 2 at x = 1 gives y = 4. They meet there.
The mistake that costs marks
The common slip is to turn the inequality the wrong way for “no real roots”. Take x² + kx + 4 = 0 with no real roots.
| Step | Wrong | Right |
|---|---|---|
| Condition | b² − 4ac > 0 | b² − 4ac < 0 |
| Working | k² − 16 > 0 | k² − 16 < 0 |
| Range of k | k < −4 or k > 4 | −4 < k < 4 |
Test a value. For k = 5, x² + 5x + 4 factorises and has roots −1 and −4, so k = 5 cannot give no real roots. It lies outside the correct range. The correct range is found by the method in solving quadratic inequalities using intervals.
Check yourself
The equation 2x² − 3x + k = 0 has two distinct real roots. Find the range of k.
Answer
Two distinct real roots means b² − 4ac > 0.
(−3)² − 4(2)(k) > 0, so 9 − 8k > 0 and 8k < 9.
So k < 9/8.
Check with k = 1: 2x² − 3x + 1 = (2x − 1)(x − 1), two distinct roots. With k = 2: 9 − 16 is negative, so there are no real roots. That is consistent.
What to study next
The discriminant tells you how many roots there are. Completing the square tells you where the turning point is. Continue with completing the square to find a turning point, then use the practice set.
If you want a teacher to check each inequality step on new questions, see online one-to-one Additional Mathematics tuition.