Find all roots of the polynomial P(x) = x ^4 + 9.?

Question

Find all roots of the polynomial P(x) = x^ 4 + 9.

a. (±√ 3 ± i √3)/2

b. (−√ 6 ± i √6)/2

c. ± √3i and ±√ 3

d. (± √6 ± i √6)/2

e. (± √3 ± i √6)/2

Answer 1

X^4=-9= 9 so x= sqrt(3) <(pi/4+kpi/2) k=0,1,2,3
for

k=0 x= (sqrt6+isqrt6)/2 and so on solution d)

Answer 2

P(x) = x ^4 + 9
0 = x^4 + 9

Doing y = x² (1):

0 = y² +9

y² = - 9
y' = 3i
y" = -3i

Substituing in (1):
y' = x'²
3i = x'²
x' =√3*√i
x' = √3*√i
x' =√3*√√-1
x' = 3² ^(1/4)*-1^1/4
x' = (3²*-1) ^(1/4)
x' = (-3²) ^(1/4)
x' = (-3) ^(1/2)
x' = √ -3
x' = √ 3*√ -1
x' = ± √3 i

y" = x"²
-3i = x"²
x" = √-3*√i
x" = √-3*√√-1
x" = √-3*1^(1/4)
x" = (-3)²^(1/4)*1^(1/4)
x" = 9^(1/4)*1^(1/4)
x" = (9*1)^(1/4)
x" = 9^(1/4)
x" = 3²^(1/4)
x" = 3^(1/2)
x" = ± √3

Answer: letter c


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Conferring:

x' = ± √3 i
x" = ± √3

x ^4 + 9 = 0

for x = √3 i

(√3 i)^4 = -9
√(3^4)* i^4 = -9
√81 * 1 = -9
√81 = -9 (OK)

for x = -√3 i

(-√3 i)^4 = -9
(-√3)^4* i^4 = -9
√(3^4)* 1= -9
√(3^4)= -9
√81 = -9 (OK)

for x = √3

(√3)^4 = -9
√(3^4)= -9
√81 = -9 (OK)

for x = -√3

(-√3)^4 = -9
√(3^4)= -9
√81 = -9 (OK)




Kisses from Brazil

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