Question

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

Then Find the other​

Answer 1

$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \large\mathfrak{\dag \: Solution :- }\:\:\:\:\:\:\:\:\:\:\:\:\:\:$

❍Given equation is :-

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

• As we know that complex roots occur in conjugate pairs.Therefore, 1 - i will be the other root of the given equation.

$\sf \red{\: \:\bigg( x - (1 + i)\bigg) \bigg(x - (1 - i) \bigg)}$

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

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

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

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

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

❍On Dividing :-

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

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

$\sf \green {[For \: details \: see \: the \:attachment ]}$

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

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

$\sf :\implies \: {x}^{2} + 4x + 1 = 0$

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

• a = 1

• b = 4

• c = 1

❍Roots of the equation is given by:-

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

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

$\sf :\implies \:x = \dfrac{ - 4 \: \pm \: \sqrt{12} }{2}$

$\sf :\implies \:x = \dfrac{ - 4 \: \pm \: 2\sqrt{3} }{2}$

$\sf :\implies \purple{ \:x \: = \: - 2 \: \pm \: \sqrt{3} }$

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

$\sf\purple{\: 1 + i},\red{ \: 1 - i},\green{ \: - 2 + \sqrt{3} } \: and\pink {\: - 2 - \sqrt{3}}$

⠀⠀⠀ ━━━━━━━━━━━━━━━━━━━⠀,

Answer 2

Answer:

hey

how are you.

Step-by-step explanation:

keep smile...

kehkasha here