Simulation · Maths · Class 11

Where the roots go when D turns negative

From the lesson Quadratics with a negative discriminant in Complex Numbers and Quadratic Equations. Change the values and watch what happens.

The idea behind it

NCERT §4.1

The roots of ax² + bx + c = 0, with a, b, c real and a ≠ 0, are x = (−b ± √D)/2a, where D = b² − 4ac.
When D < 0, √D = √(4ac − b²) i, so the roots are x = (−b ± √(4ac − b²) i)/2a. They are two non-real numbers with the same real part −b/2a and opposite imaginary parts: a conjugate pair.
Worked example: x² − 4x + 5 = 0 has D = 16 − 20 = −4, so √D = 2i and x = (4 ± 2i)/2 = 2 ± i. Check: (2 + i)² − 4(2 + i) + 5 = 3 + 4i − 8 − 4i + 5 = 0.
Worked example: x² + x + 1 = 0 has D = 1 − 4 = −3, so x = (−1 ± √3 i)/2.
Worked example: x² + 9 = 0 gives x² = −9 and x = ±3i; in general x² + k² = 0 has the roots ±ki.
The relations sum of roots = −b/a and product of roots = c/a still hold: for 2 ± i the sum is 4 and the product is (2 + i)(2 − i) = 4 + 1 = 5, matching −(−4)/1 and 5/1.
On a graph, a > 0 with D < 0 means the parabola y = ax² + bx + c stays above the x-axis. Its vertex has x = −b/2a, which is exactly the real part of the two complex roots.

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