# Quadratic equations with complex coefficients

The quadratic equations with complex coefficients that means the coefficients of the equations are not real numbers, they may be an imaginary numbers(i).Quadratic equation as,

$ax^{2} + bx + c = 0$ where a,b,c are complex numbers and a$\ne$ 0.

Roots of this quadratic equation is same as normal quadratic equation. They are given by

$\alpha = \frac{-b + \sqrt{b^{2}-4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

(i) if D = $b^{2}$ - 4ac = 0 then the roots are complex and equal.

(ii) if D = $b^{2} - 4ac \ne$ 0 then complex but unequal roots.

**Note :**Complex roots of an equation with real coefficients always occur in pairs. However, this may not be true in case of equations with complex coefficients.

For example: $x^{2}$ - 2ix - 1 = 0 has both roots equal to 'i'.

**Examples on quadratic equations with complex coefficients**

1) Solve the following quadratic equations by factorization method.(i) $x^{2}$ - 5ix - 6 = 0

**Solution :**Here a = 1, b = -5i (complex number) and c = -6

$\alpha = \frac{-b + \sqrt{b^{2}-4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

$\alpha = \frac{-(-5i) + \sqrt{(-5i)^{2}-4\times1\times(-6)}}{2\times 1}$

= $\frac{5i + \sqrt{25i^{2}+24}}{2}$

= $\frac{5i + i}{2}$

= 3i

$\beta = \frac{-(-5i) - \sqrt{(-5i)^{2}-4\times1\times(-6)}}{2\times 1}$

= $\frac{5i - \sqrt{25i^{2}+24}}{2}$

= $\frac{5i - i}{2}$

= 2i

So, the roots of the given equation are 3i and 2i.

(ii) $x^{2}$ + 10ix - 21 = 0

**Solution :**Here a = 1, b = 10i (complex number) and c = -21

$\alpha = \frac{-b + \sqrt{b^{2}-4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

$\alpha = \frac{-10i + \sqrt{(10i)^{2}-4\times1\times(-21)}}{2\times 1}$

= $\frac{-10i + \sqrt{100i^{2}+84}}{2}$

=$\frac{-10i + \sqrt{-16}}{2}$

= $\frac{-10i + 4i}{2}$

= -3i

$\beta = \frac{-10i - \sqrt{(10i)^{2}-4\times1\times(-21)}}{2\times 1}$

= $\frac{-10i - \sqrt{100i^{2}+84}}{2}$

=$\frac{-10i - \sqrt{-16}}{2}$

= $\frac{-10i - 4i}{2}$

= -7i

So, the roots of the given equation are -3i and -7i.

(iii) i$x^{2}$ - x + 12i = 0

**Solution :**Here a = i (complex number), b = -1 and c = 12i (complex number)

$\alpha = \frac{-b + \sqrt{b^{2}-4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^{2}-4ac}}{2a}$

$\alpha = \frac{-(-1) + \sqrt{1^{2}-4\times i\times(12i)}}{2\times i}$

= $\frac{1 + \sqrt{1-48i^{2}}}{2i}$

=$\frac{1 + \sqrt{1+48}}{2i}$

= $\frac{1 + 7}{2i}$

= $\frac{4}{i}$

= -4i (after rationalizing)

$\beta = \frac{-(-1) - \sqrt{1^{2}-4\times i\times(12i)}}{2\times i}$

= $\frac{1 - \sqrt{1-48i^{2}}}{2i}$

=$\frac{1 - \sqrt{1+48}}{2i}$

= $\frac{1 - 7}{2i}$

= $\frac{-3}{i}$

= 3i (after rationalizing)

So, the roots of the given equation are -4i and 3i.

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