# Complex roots of quadratic equation

The complex roots of quadratic equation are of the form $\alpha + i\beta$and $\alpha - i\beta$.These two are conjugate roots. Here $\alpha$ is a rational number where as $\sqrt{\beta}$ is an complex number.

**Proof for complex roots of quadratic equation :**

As $\alpha + i\beta$ is a root of the quadratic equation, $ax^{2}$ + bx + c = 0 so it satisfies the equation.

So put x = $\alpha + i\beta$ in $ax^{2}$ + bx + c = 0

$a(\alpha + i\beta)^{2} + b(\alpha + i\beta$) + c = 0

$a (\alpha^{2} + i^{2}\beta^{2} + 2.\alpha.i\beta) + b.\alpha + b.i\beta$ + c = 0

since i = $\sqrt{-1}$ ⇒ $i^{2}$ = -1

$a (\alpha^{2} - \beta^{2} + 2.\alpha.i\beta) + b.\alpha + b.i\beta$ + c = 0

$a.\alpha^{2} - a.\beta^{2} + 2.a.\alpha.i\beta + b.\alpha + b.i\beta$ + c = 0

$a.\alpha^{2} - a.\beta^{2} + b.\alpha + c + 2.a.\alpha.i\beta + b.i\beta$ = 0

$a.\alpha^{2} - a.\beta^{2} + b.\alpha + c + (2.a.\alpha.\beta + b.\beta)i$ = 0

$a.\alpha^{2} - a.\beta^{2} + b.\alpha + c + (2.a.\alpha.\beta + b.\beta$)i = 0 + 0.i

∴ $a.\alpha^{2} - a.\beta^{2} + b.\alpha$ + c = 0 and 2.a.$\alpha.\beta + b.\beta$= 0

Now put x = $\alpha - i\beta$ in the equation $ax^{2}$ + bx + c = 0, we get

$a(\alpha - i\beta)^{2} + b(\alpha -i\beta$) + c = 0

$a (\alpha^{2} + i^{2}\beta^{2} - 2.\alpha.i\beta) + b.\alpha - b.i\beta$ + c = 0

since i = $\sqrt{-1}$ ⇒ $i^{2}$ = -1

$a (\alpha^{2} - \beta^{2} - 2.\alpha.i\beta) + b.\alpha - b.i\beta$ + c = 0

$a.\alpha^{2} - a.\beta^{2} - 2.a.\alpha.i\beta + b.\alpha - b.i\beta$ + c = 0

$a.\alpha^{2} - a.\beta^{2} + b.\alpha + c - 2.a.\alpha.i\beta - b.i\beta$ = 0

$a.\alpha^{2} - a.\beta^{2} + b.\alpha + c - (2.a.\alpha.\beta + b.\beta)i$ = 0

$a.\alpha^{2} - a.\beta^{2} + b.\alpha + c - (2.a.\alpha.\beta + b.\beta$)i = 0 - 0.i

∴ $a.\alpha^{2} - a.\beta^{2} + b.\alpha$ + c = 0 and 2.a.$\alpha.\beta + b.\beta$= 0

Now its clear that if one root of the equation $ax^{2}$ + bx + c = 0 is $\alpha + i\beta$ then the other complex root is $\alpha - i\beta$.

Thus, ($\alpha +i\beta$) and ($\alpha - i\beta$) are imaginary conjugates roots. So quadratic equation complex roots are occur in pairs.

## Example on complex roots of quadratic equation

Find the quadratic equation with rational coefficients which has -1 + i as a root.**Solution :**As we know that, complex roots of quadratic equation are occurs in pair. If one root is - 1 + i then the other root must be -1 - i.

$\alpha$= -1 + i and $\beta$ = -1 - i

∴ sum of roots = $\alpha + \beta$ = -1 + i - 1 - i = -2

Product of roots = $\alpha . \beta$= (-1 + i) (-1 - i) = 1- $i^{2}$ = 1 + 1 = 2

The quadratic equation,

$x^{2}$ - (sum of roots).x + Product of roots = 0

$x^{2} - (\alpha + \beta).x + \alpha . \beta$ = 0

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

∴ Required quadratic equation is $x^{2}$ + 2x + 2 = 0

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