Math · Precalculus · Worked example

Solve a quadratic with complex roots

Solve x² + 2x + 5 = 0 and check one of the solutions.

x^{2} + 2 x + 5 = 0

Find the discriminant

With a = 1, b = 2 and c = 5, b² − 4ac is negative, so there are no real solutions.

2^{2} - 4 \cdot 1 \cdot 5 = - 16

√−16 = √16 · √−1 = 4i.

\sqrt{- 16} = 4 i

Divide both parts by 2.

x = \frac{- 2 \pm 4 i}{2} = - 1 \pm 2 i

x² = 1 − 4i + 4i² = −3 − 4i and 2x = −2 + 4i, so x² + 2x + 5 = (−3 − 4i) + (−2 + 4i) + 5 = 0.

(- 3 - 4 i) + (- 2 + 4 i) + 5 = 0

x = −1 + 2i or x = −1 − 2i.

Solve x² − 4x + 13 = 0.

Show the answer and explanation

The discriminant is 16 − 52 = −36, and √−36 = 6i, so x = (4 ± 6i)/2 = 2 ± 3i.

{(- 4)}^{2} - 4 \cdot 1 \cdot 13 = - 36

In Graph, y = x² + 2x + 5 stays above the x-axis, so there are no real roots. Its vertex, (−1, 4), sits above the real part of the solutions, −1.

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