Question

guys can you pls answer this? steps not required but it will be better if you can provide the steps. If it takes too long then can you please answer it quickly? I need it asap


class 11

Answer 1

AnsWer :

1.

Correct QuestioN :

Let α and β are roots of x² - x + 1 = 0, Then α⁵ + β⁵ is

SolutioN :

We have, Equation.

$\tt \dagger \: \: \: \: \: {x}^{2} - x + 1 = 0.$

☛ Compare With General Equation.

$\tt \dagger \: \: \: \: \: a{x}^{2} + bx + c = 0.$

$\tt \dagger \: \: \: \: \: a \neq0.$

Where as,

• a = 1.
• b = - 1.
• c = 1.

✎ We know, Quadratic Formula.

$\tt \dagger \: \: \: \: \: \fbox{x = \dfrac{ -b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

$\tt \longmapsto x = \dfrac{ 1 \pm \sqrt{ {( - 1)}^{2} - 4 \times 1 \times - 1 } }{2a}$

$\tt \longmapsto x = \dfrac{ 1 \pm \sqrt{ 1 - 4 } }{2}$

$\tt \longmapsto x = \dfrac{ 1 \pm \sqrt{ { - 3}} }{2}$

✎ Let,

• √- 1 = i.

$\tt \longmapsto x = \dfrac{ 1 \pm \sqrt{ {3i} } }{2}$

$\tt \mapsto \alpha = \dfrac{ 1 + \sqrt{ {3i} } }{2}$

$\tt \mapsto \beta = \dfrac{ 1 - \sqrt{ {3i} } }{2}$

$\rule{90}2$

$\tt \dagger \: \: \: \: \: Z = Cos \, \theta + iSin\, \theta$

$\tt \mapsto \alpha = \dfrac{ 1 + \sqrt{ {3i} } }{2}$

$\tt \mapsto \alpha = \dfrac{1}{2} + \dfrac{ \sqrt{ {3}i } }{2}$

$\tt \mapsto \alpha = Cos \dfrac{\pi}{3} +i Sin\dfrac{\pi}{3}$

$\tt \mapsto \beta = \dfrac{ 1 - \sqrt{ {3i} } }{2}$

$\tt \mapsto \beta = \dfrac{1}{2} - \dfrac{ \sqrt{ {3i} } }{2}$

$\tt \mapsto \beta = Cos \dfrac{\pi}{3} - i Sin\dfrac{\pi}{3}$

$\rule{90}2$

$\tt \longmapsto {e}^{i \theta}$

» Now,

• The value of theta is π / 3.

$\tt \longmapsto \alpha = {e}^{i \frac{ \pi}{ 3} }$

$\tt \longmapsto \beta = {e}^{ - i \frac{ \pi}{ 3} }$

» Now, Let Find the value of :

$\tt \longmapsto {\alpha}^{5} + { \beta }^{5} = {\bigg( {e}^{i \frac{ \pi}{3} } \bigg)}^{5} + {\bigg( {e}^{ - i \frac{ \pi}{3} } \bigg)}^{5}$

$\tt \longmapsto {\alpha}^{5} + { \beta }^{5} =cos \frac{ 5\pi}{3} + isin\frac{ 5\pi}{3} + cos \frac{ 5\pi}{3} - isin \frac{ 5\pi}{3}$

$\tt \longmapsto {\alpha}^{5} + { \beta }^{5} =2cos \frac{ 5\pi}{3}$

$\tt \longmapsto {\alpha}^{5} + { \beta }^{5} =2cos \bigg(2 \pi - \frac{ \pi}{3} \bigg)$

$\tt \longmapsto {\alpha}^{5} + { \beta }^{5} =2cos \frac{ \pi}{3}$

$\tt \longmapsto {\alpha}^{5} + { \beta }^{5} =2 \times \dfrac{1}{2}$

$\tt \longmapsto {\alpha}^{5} + { \beta }^{5} = 1.$

Answer 2

Answer:

Step-by-step explanation:

By newton law of powers

S1-1=0

S1=1

S2-s1-2=0

S2=3

S3-s2-s1=0

S3=4

S4-s3-s2=0

S4=7

S5-s4-s3=0

S5=11