Question

If alpha and beta are the zeroes of the equation 2x^2+10x-12=0.findnthe value of alpha ^2+beta^2 and alpha^3+beta^3

Answer 1

Answer:

$let \: \\ \alpha \: and \: \beta \: are \: zeroes \: of \: giv en\\ \\ polynomial \\ then \\ \alpha + \beta = \frac{ - b}{a} \\ \alpha + \beta = \frac{ - (10)}{2} \\ \alpha + \beta = - 5 \\ \alpha \beta = \frac{c}{a} \\ \alpha \beta = \frac{ - 12}{2 } \\ \alpha \beta = - 6$

${ \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ { \alpha }^{2} + { \beta }^{2} = {( - 5)}^{2} - 2( - 6) \\ { \alpha }^{2} + { \beta }^{2} = 25 + 12 = 37$

${\alpha }^{3} + { \beta }^{3} = ( \alpha + \beta )( { \alpha }^{2} + { \beta }^{2} - \alpha \beta ) \\ { \alpha }^{3} + { \beta }^{3} = ( - 5)(37 - ( - 6) \\ { \alpha }^{3} + { \beta }^{3} = ( - 5)(37 + 6) = - 215$

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Answer 2

Step-by-step explanation:

Given Equation is 2x² + 10x - 12 = 0

a = 2, b = 10, c = -12.

Sum of zeroes:

α + β = -b/a

        = -10/2

        = -5

Product of zeroes = c/a

                              = -12/2

                              = -6

Now,

α² + β² = (α + β)² - 2αβ

 

            = (-5)² - 2(-6)

            = 25 + 12

            = 37

(ii)

α³ + β³ = (α + β)(α² - αβ + β²)

           = (-5)(37 + 6)

           = -215

Hope it helps!