Question

if one root of quadratic equation is 4+3i then find quadratic equation


Please help I'll mark brainliest​

Answer 1

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

Given that,

• One root of the quadratic equation is 4 + 3i

We know,

• Complex roots occur in conjugate pairs

So,

• Other root of the quadratic equation is 4 - 3i

So, Let we assume that

$\rm :\longmapsto\: \alpha = 4 + 3i$

and

$\rm :\longmapsto\: \beta = 4 - 3i$

So,

$\rm :\longmapsto\: \alpha + \beta = 4 + 3i + 4 - 3i$

$\rm :\longmapsto\: \alpha + \beta = 8$

Also,

$\rm :\longmapsto\: \alpha \beta = (4 + 3i)(4 - 3i)$

$\rm :\longmapsto\: \alpha \beta = {4}^{2} - {(3i)}^{2}$

$\rm :\longmapsto\: \alpha \beta = 16 - 9 {i}^{2}$

$\rm :\longmapsto\: \alpha \beta = 16 - 9( - 1)$

$\rm :\longmapsto\: \alpha \beta = 16 + 9$

$\rm :\longmapsto\: \alpha \beta = 25$

Thus,

The required Quadratic equation is

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

So, on substituting the values, we get

$\bf :\longmapsto\: {x}^{2} - 8x + 25 = 0$

Alternative Method :-

Given that,

One root of the quadratic equation is 4 + 3i

$\rm \implies\:x = 4 + 3i$

$\rm \implies\:x - 4 = 3i$

On squaring both sides, we get

$\rm :\longmapsto\: {(x - 4)}^{2} = {(3i)}^{2}$

$\rm :\longmapsto\: {x}^{2} + 16 - 8x = 9 {i}^{2}$

$\rm :\longmapsto\: {x}^{2} + 16 - 8x = 9( - 1)$

$\rm :\longmapsto\: {x}^{2} + 16 - 8x = - 9$

$\rm :\longmapsto\: {x}^{2} - 8x + 16 + 9 = 0$

$\bf :\longmapsto\: {x}^{2} - 8x + 25 = 0$

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Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac