Question

x=-5+4i then xPower 4 +9x cube+35x^-x+4 =

Answer 1

Step-by-step explanation:

GIVEN :-

$\sf \: x = - 5 + 4i$

TO FIND :-

$\sf \: {x}^{4} + 9 {x}^{3} + 35 {x}^{2} - x + 4$

SOLUTION :-

$\sf \: It \: \: is \: \: given \: \: that -$

$\sf \: x = - 5 + 4i$

$\implies \sf \: x + 5= 4i \: \: \: \: \: \: ......(i)$

$\sf \: Squaring \: \: both \: \: sides -$

$\implies \sf \: (x + 5)^{2} =( 4i)^{2}$

$\sf \: As \: \: we \: \: know \: \: that -$

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

$\implies \sf \: x ^{2} + 10x + 25= - 16$

$\implies \sf \: x ^{2} + 10x + 25 + 16 = 0$

$\implies \sf \: x ^{2} + 10x + 41 = 0 \: \: \: \: \: \: \: \: \: .....(ii)$

$\large \sf \: Now \: \: according \: \: to \: \: the \: \: Question -$

$\sf \: {x}^{4} + 9 {x}^{3} + 35 {x}^{2} - x + 4$

$= \sf \: {x}^{2} ( {x}^{2} + 9x + 35) - x + 4$

$\sf \: = {x }^{2} ( {x}^{2} + (10x - x )+ (41 - 6)) - x + 4$

$\sf \: = {x}^{2} ( {x}^{2} + 10x + 41) - {x}^{3} - 6 {x}^{2} - x + 4$

$\sf \: = {x}^{2} \times 0 - x( {x}^{2} + 6x + 1) + 4$

$\sf \: = 0 - x( {x}^{2} + (10x - 4x) + (41 - 40)) + 4$

$\sf = - x( {x}^{2} + 10x + 41) + 4 {x}^{2} + 40x + 4$

$\sf \: = - x( {x}^{2} + 10x + 41) + 4( {x}^{2} + 10x + 1)$

$\sf \: = - x(0) + 4(0)$

$\sf = 0$

$\sf \: \orange{ Hence \: Required \: ans \: }$

$\green{ \sf \: {x}^{4} + 9 {x}^{3} + 35 {x}^{2} - x + 4 = \orange{0}}$