Slove the equation x^4+2x^3-5x^2+6x+2=0 given that 1+i is one of its r

1.	Slove the equation x^4+2x^3-5x^2+6x+2=0 given that 1+i is one of its roots
Answer» $\large\underline{\sf{<klux>GIVEN</klux> - }}$ $\sf \: An \: equation : {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0\: having \: one \: root \: 1 + i$ $\begin{gathered}\begin{gathered}\bf \: To\: find - \begin{cases} &\sf{remaining \: roots \: of \: equation} \end{cases}\end{gathered}\end{gathered}$ $\large\underline{\bold{Solution-}}$ Given equation is $\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0\:having \: one \: root \: 1 + i$ We know, complex roots occur in conjugate pairs. So, 1 - i is the other root of the given equation. So, $\sf \: \:\bigg( x - (1 + i)\bigg) \bigg(x - (1 - i) \bigg) \: is \: a \: factor \: of \:given \: equation$ $\sf \: = \: (x - 1 - i)(x - 1 + i)$ $\sf \: = {(x - 1)}^{2} - {i}^{2}$ $\sf \: = {x}^{2} + 1 - 2x - ( - 1) \: \: \: \: \: ( \bf \because \: {i}^{2} = - 1)$ $\sf \: = {x}^{2} - 2x + 2$ So, it means $\sf \: {x}^{2} - 2x + 2 \: is \: a \: factor \: of \: given \: equation.$ So on dividing $\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 \: by \: {x}^{2} - 2x + 2 \: we \: get \:$ $\sf \: {x}^{2} + 4x + 1$ Therefore, $\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = ( {x}^{2} - 2x + 2)( {x}^{2} + 4x + 1)$ Hence, $\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0\:$ $\rm :\implies\: ({x}^{2} - 2x + 2)( {x}^{2} + 4x + 1) = 0$ As we know that roots of the equation x² - 2x + 2 = 0 are 1 + i and 1 - i, So, we have to find the roots of the equation $\rm :\longmapsto\: {x}^{2} + 4x + 1 = 0$ We know, the roots of quadratic is given by Quadratic formula, So, On comparing it with ax² + bx + C = 0, we have a = 1 b = 4 c = 1 So, Roots of the equation is given by $\rm :\longmapsto\:\sf \: x \: = \: \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$ $\rm :\longmapsto\:x \: = \: \dfrac{ - 4 \: \pm \: \sqrt{ {4}^{2} - 4 \times 1 \times 1} }{2 \times 1}$ $\rm :\longmapsto\:x = \dfrac{ - 4 \: \pm \: \sqrt{12} }{2}$ $\rm :\longmapsto\:x = \dfrac{ - 4 \: \pm \: 2\sqrt{3} }{2}$ $\rm :\longmapsto\:x \: = \: - 2 \: \pm \: \sqrt{3}$ Hence, $\bf \: The \: roots \: of \: \sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0\: \: are$ $\bf \: 1 + i, \: 1 - i, \: - 2 + \sqrt{3} \: and \: - 2 - \sqrt{3}$ ADDITIONAL Information :- Let US consider a BI - quadratic equation $\sf \: {ax}^{4} + {bx}^{3} + {cx}^{2} + dx + e = 0 \:$ having roots p, q, r and s, then $\sf \: p + q + r + s = - \dfrac{b}{a}$ $\sf \: pq + pr + ps + qr + qs + sr = \dfrac{c}{a}$ $\sf \: pqr + pqs + qrs + rsp = - \dfrac{d}{a}$ $\sf \: pqrs = \dfrac{e}{a}$

Slove the equation x^4+2x^3-5x^2+6x+2=0 given that 1+i is one of its roots

Answer»

$\large\underline{\sf{<klux>GIVEN</klux> - }}$

$\sf \: An \: equation : {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0\: having \: one \: root \: 1 + i$

$\begin{gathered}\begin{gathered}\bf \: To\: find - \begin{cases} &\sf{remaining \: roots \: of \: equation} \end{cases}\end{gathered}\end{gathered}$

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

Given equation is

$\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0\:having \: one \: root \: 1 + i$

We know, complex roots occur in conjugate pairs.

So, 1 - i is the other root of the given equation.

So,

$\sf \: \:\bigg( x - (1 + i)\bigg) \bigg(x - (1 - i) \bigg) \: is \: a \: factor \: of \:given \: equation$

$\sf \: = \: (x - 1 - i)(x - 1 + i)$

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

$\sf \: = {x}^{2} + 1 - 2x - ( - 1) \: \: \: \: \: ( \bf \because \: {i}^{2} = - 1)$

$\sf \: = {x}^{2} - 2x + 2$

So,

it means

$\sf \: {x}^{2} - 2x + 2 \: is \: a \: factor \: of \: given \: equation.$

So on dividing

$\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 \: by \: {x}^{2} - 2x + 2 \: we \: get \:$

$\sf \: {x}^{2} + 4x + 1$

Therefore,

$\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = ( {x}^{2} - 2x + 2)( {x}^{2} + 4x + 1)$

Hence,

$\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0\:$

$\rm :\implies\: ({x}^{2} - 2x + 2)( {x}^{2} + 4x + 1) = 0$

As we know that roots of the equation x² - 2x + 2 = 0 are 1 + i and 1 - i,

So, we have to find the roots of the equation

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

We know, the roots of quadratic is given by Quadratic formula,

So,

On comparing it with ax² + bx + C = 0, we have

a = 1

b = 4

c = 1

So,

Roots of the equation is given by

$\rm :\longmapsto\:\sf \: x \: = \: \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

$\rm :\longmapsto\:x \: = \: \dfrac{ - 4 \: \pm \: \sqrt{ {4}^{2} - 4 \times 1 \times 1} }{2 \times 1}$

$\rm :\longmapsto\:x = \dfrac{ - 4 \: \pm \: \sqrt{12} }{2}$

$\rm :\longmapsto\:x = \dfrac{ - 4 \: \pm \: 2\sqrt{3} }{2}$

$\rm :\longmapsto\:x \: = \: - 2 \: \pm \: \sqrt{3}$

Hence,

$\bf \: The \: roots \: of \: \sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0\: \: are$

$\bf \: 1 + i, \: 1 - i, \: - 2 + \sqrt{3} \: and \: - 2 - \sqrt{3}$

ADDITIONAL Information :-

Let US consider a BI - quadratic equation

$\sf \: {ax}^{4} + {bx}^{3} + {cx}^{2} + dx + e = 0 \:$

having roots p, q, r and s, then

$\sf \: p + q + r + s = - \dfrac{b}{a}$

$\sf \: pq + pr + ps + qr + qs + sr = \dfrac{c}{a}$

$\sf \: pqr + pqs + qrs + rsp = - \dfrac{d}{a}$

$\sf \: pqrs = \dfrac{e}{a}$