Question

If alpha and beta are the roots of equation x2+3x+2=0 then alpha5+ beta5=

Answer 1

Step-by-step explanation:

Given -

α and β are zeroes of polynomial x² + 3x + 2 = 0

To Find -

α^5 + β^5

In x² + 3x + 2

here,

a = 1

b = 3

c = 2

Now,

Using Quadratic formula -

• x = -b ± √b² - 4ac/2a

» -(3) ± √(3)² - 4×1×2/2(1)

» -3 ± √9 - 8/2

» -3 ± √1/2

» -3 ± 1/2

Zeroes are -

• x = -3 - 1/2 = -4/2 = -2

and

• x = -3 + 1/2 = -2/2 = -1

Let α = -2 and β = -1

Then,

The value of α^5 + β^5 is

» (-2)^5 + (-1)^5

» -32 + (-1)

» -32 - 1

• » -33

Hence,

The value of α^5 + β^5 is -33

Answer 2

$\bold{\huge{\underline{\underline{\rm{ Given :}}}}}$

$\alpha \: and \: \beta \: are \: the \: roots \: of \: \\ equation \: {x}^{2} + 3x + 2 = 0$

$\bold{\huge{\underline{\underline{\rm{ To\:Find :}}}}}$

${ \alpha }^{5} + { \beta }^{5}$

$\rule{200}{1}$

$\huge{\underline{\underline{\purple{♡Solution→}}}}$

First find the roots of Given equation -

${x}^{2} + 3x + 2 = 0 \\ {x}^{2} + 2x + x + 2 = 0 \\ x(x + 2) + 1(x + 2) = 0 \\ (x + 1)(x + 2) = 0$

$\rule{200}{1}$

First Root :-

$x + 1 = 0 \\ x = - 1$

We can say that :-

$\alpha = - 1$

$\rule{200}{1}$

Second Root :-

$x + 2 = 0 \\ x = - 2$

We can say that :-

$\beta = - 2$

$\rule{200}{1}$

Now the question is -

${ \alpha }^{5} + { \beta }^{5} \\ = ( { - 1)}^{5} + ( { - 2)}^{5} \\ = - 1 +( ( - 2) \times ( - 2) \times ( - 2) \times ( - 2) \times ( - 2)) \\ = - 1 - 32 \\ = - 33$

$\rule{200}{1}$

Hence :-

$\boxed{ { \alpha }^{5} + { \beta }^{5} = - 33}$

$\rule{200}{1}$

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