Question

Form a Quadratic equation with real coefficients if one root is -1+2i

Answer 1

$\large\underline{\sf{Solution-}}$

Given that one root of the quadratic equation is- 1 + 2i.

Let assume that,

$\rm :\longmapsto\:x = - 1 + 2i$

can be rewritten as

$\rm :\longmapsto\:x + 1= 2i$

On squaring both sides, we get

$\rm :\longmapsto\: {(x + 1)}^{2} = {(2i)}^{2}$

$\rm :\longmapsto\: {x}^{2} + 1 + 2x = 4 {i}^{2}$

$\rm :\longmapsto\: {x}^{2} + 1 + 2x = - 4 \: \: \: \: \: \: \: \: \{ \because \: {i}^{2} = - 1 \}$

$\rm :\longmapsto\: {x}^{2} + 1 + 2x + 4 =0$

$\bf\implies \:\boxed{ \tt{ \: {x}^{2} + 2x + 5 = 0 \: }}$

is the required Quadratic equation.

Alternative Method :-

Given that one root of the quadratic equation is- 1 + 2i.

We know, complex roots occur in conjugate pairs.

So, roots of the equation are

$\rm :\longmapsto\: \alpha = - 1 + 2i$

and

$\rm :\longmapsto\: \beta = - 1 - 2i$

Now,

Consider,

$\rm :\longmapsto\: \alpha + \beta$

$\rm \: = \: - 1 + 2i - 1 - 2i$

$\rm \: = \: - \: 2$

$\rm \implies\:\boxed{ \tt{ \: \alpha + \beta = - \: 2 \: }}$

Now, Consider

$\rm :\longmapsto\: \alpha \beta$

$\rm \: = \: ( - 1 + 2i)( - 1 - 2i)$

$\rm \: = \: {( - 1)}^{2} - {(2i)}^{2}$

$\rm \: = \:1 - {4i}^{2}$

$\rm \: = \:1 - 4 \times ( - 1)$

$\rm \: = \:1 + 4$

$\rm \: = \:5$

$\rm \implies\:\boxed{ \tt{ \: \alpha \beta = \: 5 \: }}$

Now, Required Quadratic equation is given by

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0 \: }}}$

So, on substituting the values, we get

$\rm :\longmapsto\: {x}^{2} - ( - 2)x + 5 = 0$

$\bf\implies \:\boxed{ \tt{ \: {x}^{2} + 2x + 5 = 0 \: }}$

is the required Quadratic equation