Question

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If α and β are the roots of the equation x2 + 2x – 2 = 0, then what is the equation whose roots are α5 and β5?

Answer 1

x^2 + 2x - 2 = 0

if a and b root of any equation then :-

x^2 - ( a+b)x + ab = 0

then

a+b = -2
ab = -2



and now we find the value of

a^5 + b^5 = ??

then


( a+ b )^5 = a^5 + 5a^4.b + 10a^3. b^2 + 10a^2.b^3 + 5a.b^4 + b^5


(a+b)^5 - 5a^4.b - 5a.b^4 - 10a^3.b^2 - 10 a^2.b^3 = a^5 + b^5

(-2)^5 - 5(-2)(a^3 + b^3) - 10(-2)^2 . ( -2) = a^5 + b^5


-32 + 10 {(a+b)^3 - 3ab} + 80 = a^5 + b^5

-32 + 10 { (-2)^3 - 3(-2) } + 80 = a^5 + b^5

-32 + 10{ -8 + 6} + 80 = a^5 + b^5

48 -20 = a^5 + b^5

28 = a^5 + b^5

Answer 2

${x}^{2} + 2x - 2 = 0 \\ \alpha + \beta = - \frac{b}{a} = - \frac{2}{1} = - 2 \\ \alpha \beta = \frac{c}{a} = \frac{ - 2}{1} = - 2 \\{ ( \alpha + \beta )}^{2} = { \alpha}^{2} + { \beta }^{2} + 2 \alpha \beta \\ ( { - 2})^{2} = { \alpha}^{2} + { \beta }^{2} + 2( - 2) \\ { \alpha}^{2} + { \beta }^{2} = 4 + 4 = 8...............(1) \\ now \: on\: cubing \: we \: get \\ ({ \alpha + \beta })^{3} = { \alpha}^{3} + { \beta }^{3} + 3 \alpha \beta (\alpha + \beta ) \\ {( - 2)}^{3} = { \alpha}^{3} + { \beta }^{3} + 3( - 2)( - 2) \\ - 8 = { \alpha}^{3} + { \beta }^{3} + 12 \\ { \alpha}^{3} + { \beta }^{3} = - 8 -12 = - 20.......(2) \\ multiplying \: (1) \: and \: (2) \\ ({ \alpha}^{2} + { \beta }^{2} ) ({ \alpha}^{3} + { \beta }^{3} ) = { \alpha}^{5} + { \alpha}^{2} { \beta }^{3} + { \beta }^{2} { \alpha}^{3} + { \beta }^{5} \\ (8)( - 20) = {\alpha}^{5} + { \beta }^{5} + { \alpha}^{2} { \beta }^{2} ( \alpha + \beta ) \\ - 160 = {\alpha}^{5} + { \beta }^{5} + {( - 2)}^{2} ( - 2) \\ - 160 = {\alpha}^{5} + { \beta }^{5} - 8 \\ {\alpha}^{5} + { \beta }^{5} = - 160 + 8 = - 152..........(3) \\ {\alpha}^{5} { \beta }^{5} = {( - 2)}^{5} = - 32 \\ the \: required \: equation \: whose \: roots \: are \: { \alpha }^{5} and \: { \beta }^{5} is \\ {x}^{2} - ( - 152)x + ( - 32) = 0 \\ {x}^{2} + 152x - 32 = 0$