Question

If alpha and beta are the roots of polynomial x2 + 3x + 2 then
find alpha power 4 +beta power 4

Answer 1

heya..!!!!!


x² + 3x + 2

=> x² + 2x + x + 2

=> x(x+2) + (x+2)

=> (x+1)(x+2)

=> x = -1 = alpha

=> (alpha)⁴ = (-1)⁴ = 1

AND,

x = -2 = beta

=> (beta)⁴ = (-2)⁴ = (-2)(-2)(-2)(-2)

=> 16

alpha ⁴ + beta ⁴ = 1 + 16 = 17

Answer 2

Answer:

$\huge{\pink{\boxed{\green{\sf{\therefore \alpha^{4}+\beta^{4}=17}}}}}$

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\: \: \: \: {\orange{given}} \\ {\pink{ \boxed {\green{ {x}^{2} + 3x + 2 = 0 \: has \: two \: roots \: \alpha \: and \: \beta }}}} \\ \\ \: \: { \blue{to \: find}} \\ {\purple{\boxed{\red{{ \alpha }^{4} + { \beta }^{4} = }}}}$

☆ According to given question:

☆ find roots of the eqn:

☆ Solve the eqn by middle term spliting:

$\to {x}^{2} + 3x + 2 = 0 \\ \to {x}^{2} + 2x + x + 2 = 0 \\ \to x(x + 2) + 1(x + 2) = 0 \\ \to(x + 2)(x + 1) = 0 \\ \\ \to \alpha = x + 2 = 0 \\ { \boxed{\to \alpha = x = - 2 }}\\ \\ \to \beta = x + 1 = 0 \\ { \boxed {\to \beta = x = - 1}} \\ \\ \to { \alpha }^{4} + { \beta }^{4} \\ \to( { -2 })^{4} + ({ - 1})^{2} \\ \to 16 + 1 \\ { \pink{ \boxed{ \green{\therefore { \alpha }^{4} + { \beta }^{4} = 17}}}}$

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