Question

15. If alpha and Beta be two zeroes of the quadratic
polynomial p(x) = 2x²-3x+7 evaluate

1) 1/ alpha +1/beta

2) alpha ²+ beta²

3)alpha³+beta³
​

Answer 1

Answer:

α  and β  are    zeros / roots of   2 x² - 3 x + 7

        So  the sum of the roots = α + β =  - (-3/2) = 1.5 and   

              the product of the roots = α β =  7 / 2  = 3.5

Answer 2

Answer:-

$\bf \: as \: \: \alpha \: and \: \: \beta \: \: are \: zeroes \: of \: polynomial \: \: \\ \\ \\ \tt \: p( x) = {2x}^{2} - 3x + 7 \\ \\ \\ \tt \therefore \: \alpha + \beta = \frac{ - ( - 3)}{2} = \frac{3}{2} \: \: and \: \alpha \beta = \frac{7}{2}$

$\tt \: 1) \tt \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ \beta + \alpha }{ \alpha \beta } = \frac{3}{7}$

$\tt \: 2) {( \alpha + \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta$

$\tt \implies \: {( \frac{3}{2} )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \times \frac{7}{2}$

$\tt \implies \: { \alpha }^{2} + { \beta }^{2} = \frac{9}{4} - 7 = \frac{9 - 28}{4} = - \frac{19}{4}$

$\tt 3) {( \alpha + \beta )}^{3} = { \alpha }^{3} + { \beta }^{3} + 3 \alpha \beta ( \alpha + \beta )$

$\tt \implies \: {( \frac{3}{2} )}^{2} = { \alpha }^{3} + { \beta }^{3} + 3 \times \frac{7}{2} \times \frac{3}{2}$

$\tt \implies \: { \alpha }^{3} + { \beta }^{3} = \frac{27}{8} - \frac{63}{4} = \frac{27 - 126}{8} = - \frac{99}{8}$