SATHEE: Theory of Equations 1 Question 11

Theory of Equations 1 Question 11

12. Let $α$ and $β$ be two roots of the equation $x^{2} + 2 x + 2 = 0$ , then $α^{15} + β^{15}$ is equal to

(2019 Main, 9 Jan I)

(a) 256

$t$ (b) 512

(d) -512

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Answer:

Correct Answer: 12. (c)

Solution:

We have, $x^{2} + 2 x + 2 = 0$

$\Rightarrow x = \frac{- 2 \pm \sqrt{4 - 8}}{2} [∵$ roots of $a x^{2} + b x + c = 0$ are given by $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}]$

$\Rightarrow x = - 1 \pm i$

Let $α = - 1 + i$ and $β = - 1 - i$ .

Then, $α^{15} + β^{15} = (- 1 + i)^{15} + (- 1 - i)^{15}$

$\begin{aligned} = - [(1 - i)^{15} + (1 + i)^{15}] \\ = - \sqrt{2} \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} + \sqrt{2} \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \\ = - \sqrt{2} \cos \frac{π}{4} - i \sin \frac{π}{4} \\ + \sqrt{2} \cos \frac{π}{4} + i \sin \frac{π}{4} \\ = - (\sqrt{2})^{15} \cos \frac{15 π}{4} - i \sin \frac{15 π}{4} + \cos \frac{15 π}{4} + i \sin \frac{15 π}{4} \end{aligned}$

[using De’ Moivre’s theorem $(\cos θ \pm i \sin θ)^{n} = \cos n θ \pm i \sin n θ, n \in Z]$

$= - (\sqrt{2})^{15} 2 \cos \frac{15 π}{4} = - (\sqrt{2})^{15} 2 \times \frac{1}{\sqrt{2}}$

$∵ \cos \frac{15 π}{4} = \cos 4 π - \frac{π}{4} = \cos \frac{π}{4} = \frac{1}{\sqrt{2}}$

$= - (\sqrt{2})^{16} = - 2^{8} = - 256$ .

Alternate Method

$\begin{aligned} α^{15} + β^{15} & = (- 1 + i)^{15} + (- 1 - i)^{15} \\ = - [(1 - i)^{15} + (1 + i)^{15}] \\ = - \frac{(1 - i)^{16}}{1 - i} + \frac{(1 + i)^{16}}{1 + i} \\ = - \frac{{[(1 - i)^{2}]}^{8}}{1 - i} + \frac{{[(1 + i)^{2}]}^{8}}{1 + i} \end{aligned}$

$\begin{aligned} = - \frac{{[1 + i^{2} - 2 i]}^{8}}{1 - i} + \frac{{[1 + i^{2} + 2 i]}^{8}}{1 + i} \\ = - \frac{(- 2 i)^{8}}{1 - i} + \frac{(2 i)^{8}}{1 + i} \\ = - 2^{8} \frac{1}{1 - i} + \frac{1}{1 + i} [∵ i^{4 n} = 1, n \in Z] \\ = - 256 \frac{2}{1 - (i)^{2}} = - 256 \frac{2}{2} = - 256 \end{aligned}$

Theory of Equations 1 Question 11

12. Let $α$ and $β$ be two roots of the equation $x^{2} + 2 x + 2 = 0$ , then $α^{15} + β^{15}$ is equal to

Answer:

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Theory of Equations 1 Question 11

12. Let α and β be two roots of the equation x2+2x+2=0, then α15+β15 is equal to

Answer:

Alternate Method

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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12. Let $α$ and $β$ be two roots of the equation $x^{2} + 2 x + 2 = 0$ , then $α^{15} + β^{15}$ is equal to