Question

$Let \: a \: and \: b \: be \: two \: \\ roots \: of \: the \: equation \ \\ {x}^{2} +2x+2=0, \: then \: \\ {a}^{15} + {b}^{15} \: is \: equal \: to$
answer with solution plz
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Answer 1

Answer:

-256

Step-by-step explanation:

Given,

a and b are the two roots of the equation.

Equation is x² + 2x + 2 = 0

=> x² + 2x + 1 + 1 = 0

=> x² + 2x + 1 = -1

=> (x + 1)² = -1

=> (x + 1) = ± i

=> x = -1 ± i

∴ Roots are -1 + i and -1 - i

Thus,

a¹⁵ + b¹⁵

=> (-1 + i)¹⁵ + (-1 - i)¹⁵

=> -128 - 128i + -128 + 128i

=> -256

Hope it helps!

Answer 2

Step-by-step explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}\mid}}}}$

$\huge \bold{GIVEN:- }\\ \\ {x}^{2} + 2x + 2 = 0 \huge{ [ }^{a}_b$

$\huge\bold{TO FIND:-} \\ \\ {a}^{15} + {b}^{15}$

${x}^{2} + 2x + 2 = 0 \\ \\ {x}^{2} + 2x + 1 + 1 = 0 \\ \\ ( {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab ) \\ \\ \implies {(x + 1)}^{2} = - 1 \\ \\ We \: \: know \: \: \sqrt{ - 1} = \iota \\ \\ \implies x + 1 = \pm \: \iota \\ \\ \implies x = - 1 \pm \: \iota \\ \\ \boxed{a = - 1 + \iota \: \: \: \: \: \: \: b = - 1 - \iota}$

$Comparing \: \: with \: \: a + \iota b \\ \\ a = - 1 \: \: \: b = \pm 1$

$r = \sqrt{ {a}^{2} + {b}^{2} } = \sqrt{1 + 1} = \sqrt{2}$

$Using \: \: complex \: \: numbers.. \\ \\ r [sin(\theta) + cos(\theta)] \\ \\ \implies a = -1 + \iota = \sqrt{2} \times [ \cos (\frac{3 \pi}{4} ) + \iota \sin(\frac{3 \pi}{4})] \\ \\ \implies b = -1 - \iota = \sqrt{2} \times [\cos (\frac{3 \pi}{4} - \iota \sin(\frac{3 \pi}{4} )$

$\implies {a}^{15} = {\sqrt{2}}^{15} \times { \cos (\frac{3 \pi}{4} ) + \iota \sin (\frac{3 \pi}{4}) }^{15} \\ \\ \implies {b}^{15} = {\sqrt{2}}^{15} \times { \cos (\frac{3 \pi}{4} ) + \iota \sin(\frac{3 \pi}{4}) }^{15}$ $\implies {a}^{15} + {b}^{15} = {\sqrt{2}}^{15} \times [ {\cos (\frac{3 \pi}{4} \cancel{+ \iota \sin(\frac{3 \pi}{4})} }^{15} + \\ \\ { \cos (\frac{3 \pi}{4} ) \cancel{- \iota \sin(\frac{3 \pi}{4})} }^{15}$

$By \: \: De \: \: Morgan's \: \: Theorem, \\ \\ \implies {a}^{15} + {b}^{15} = {\sqrt{2}}^{15} \times [ 2 \cos(\frac{(3 \times 15) \pi}{4} \\ \\ \implies {a}^{15} + {b}^{15} = {\sqrt{2}}^{15} \times 2 \times \frac{ -1 }{\sqrt{2}} \\ \\ \implies {a}^{15} + {b}^{15} = 128 \times \cancel{\sqrt{2}} \times 2 \times \frac{-1}{\cancel{\sqrt{2}}} \\ \\ \huge\boxed{\implies - 256 }$