Find the quadratic equation whose roots is (3±i√5)/2

1.	Find the quadratic equation whose roots is (3±i√5)/2
Answer» $\large\underline{\sf{Solution-}}$ Given that roots of the QUADRATIC equation are $\rm :\longmapsto\:\dfrac{3 + \sqrt{5} i}{2}$ and $\rm :\longmapsto\:\dfrac{3 - \sqrt{5} i}{2}$ Let assume that, $\rm :\longmapsto\: \alpha = \dfrac{3 + \sqrt{5} i}{2}$ and $\rm :\longmapsto\: \beta = \dfrac{3 - \sqrt{5} i}{2}$ Now, Consider $\red{\rm :\longmapsto\: \alpha + \beta }$ $\rm \: = \: \: \dfrac{3 + \sqrt{5} i}{2} + \dfrac{3 - \sqrt{5} i}{2}$ $\rm \: = \: \: \dfrac{3 + \sqrt{5} i + 3 - \sqrt{5}i }{2}$ $\rm \: = \: \: \dfrac{6}{2}$ $\rm \: = \: \: 3$ $\bf\implies \: \alpha + \beta = 3$ Now, Consider $\red{\rm :\longmapsto\: \alpha \beta }$ $\rm \: = \: \: \dfrac{3 + \sqrt{5} i}{2} \: \: \times \: \: \dfrac{3 - \sqrt{5} i}{2}$ We know, $\boxed{ \rm{ (x + y)(x - y) = {x}^{2} - {y}^{2}}}$ So, using this identity, we get $\rm \: = \: \: \dfrac{ {3}^{2} - (\sqrt{5} i) {}^{2} }{4}$ $\rm \: = \: \: \dfrac{9 - 5 {i}^{2} }{4}$ $\rm \: = \: \: \dfrac{9 + 5}{4} \: \: \: \: \: \: \: \: \: \{ \because \: {i}^{2} = - \: 1 \}$ $\rm \: = \: \: \dfrac{14}{4}$ $\rm \: = \: \: \dfrac{7}{2}$ $\bf\implies \: \alpha \beta \: = \: \: \dfrac{7}{2}$ Now, Required Quadratic equation is given by, $\red{\rm :\longmapsto\: {x}^{2} - ( \alpha + \beta ) + \alpha \beta = 0}$ On SUBSTITUTING the VALUES, we have $\rm :\longmapsto\: {x}^{2} - 3x + \dfrac{7}{2} = 0$ can be REWRITTEN as $\rm :\longmapsto\: {2x}^{2} - 6x + 7 = 0$ 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

Answer»

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

Given that roots of the QUADRATIC equation are

$\rm :\longmapsto\:\dfrac{3 + \sqrt{5} i}{2}$

and

$\rm :\longmapsto\:\dfrac{3 - \sqrt{5} i}{2}$

Let assume that,

$\rm :\longmapsto\: \alpha = \dfrac{3 + \sqrt{5} i}{2}$

and

$\rm :\longmapsto\: \beta = \dfrac{3 - \sqrt{5} i}{2}$

Now, Consider

$\red{\rm :\longmapsto\: \alpha + \beta }$

$\rm \: = \: \: \dfrac{3 + \sqrt{5} i}{2} + \dfrac{3 - \sqrt{5} i}{2}$

$\rm \: = \: \: \dfrac{3 + \sqrt{5} i + 3 - \sqrt{5}i }{2}$

$\rm \: = \: \: \dfrac{6}{2}$

$\rm \: = \: \: 3$

$\bf\implies \: \alpha + \beta = 3$

Now, Consider

$\red{\rm :\longmapsto\: \alpha \beta }$

$\rm \: = \: \: \dfrac{3 + \sqrt{5} i}{2} \: \: \times \: \: \dfrac{3 - \sqrt{5} i}{2}$

We know,

$\boxed{ \rm{ (x + y)(x - y) = {x}^{2} - {y}^{2}}}$

So, using this identity, we get

$\rm \: = \: \: \dfrac{ {3}^{2} - (\sqrt{5} i) {}^{2} }{4}$

$\rm \: = \: \: \dfrac{9 - 5 {i}^{2} }{4}$

$\rm \: = \: \: \dfrac{9 + 5}{4} \: \: \: \: \: \: \: \: \: \{ \because \: {i}^{2} = - \: 1 \}$

$\rm \: = \: \: \dfrac{14}{4}$

$\rm \: = \: \: \dfrac{7}{2}$

$\bf\implies \: \alpha \beta \: = \: \: \dfrac{7}{2}$

Now, Required Quadratic equation is given by,

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

On SUBSTITUTING the VALUES, we have

$\rm :\longmapsto\: {x}^{2} - 3x + \dfrac{7}{2} = 0$

can be REWRITTEN as

$\rm :\longmapsto\: {2x}^{2} - 6x + 7 = 0$

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.