Question

If a and B zeros of the quadratic polynomial x²-3x+5,then evaluate : I) a² + B² ii) a³ + ß³​

Answer 1

Answer:

${ \alpha }^{2} + { \beta }^{2} = - 1. \\ \\ { \alpha }^{3} + { \beta }^{3} = - 18.$

Step-by-step explanation:

${x}^{2} - 3x + 5 = 0 \\ \\ \alpha + \beta = 3 \\ \\ \\ \alpha \beta = 5 \\ \\ (i) \: \: \: \: \: \: \: \: \: \: \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta \\ \\ = {3}^{2} - 2 \times 5 \\ \\ = 9 - 10 \\ \\ = - 1. \\ \\ \\ \\ (ii) \: \: \: \: \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta ) \\ \\ = {3}^{3} - 3(5)(3) \\ \\ = 27 - 45 \\ \\ = - 18.$

Answer 2

Answer:

�

2

+

�

2

=

−

1.

�

3

+

�

3

=

−

18.

α

2

+β

2

=−1.

α

3

+β

3

=−18.

Step-by-step explanation:

�

2

−

3

�

+

5

=

0

�

+

�

=

3

�

�

=

5

(

�

)

�

2

+

�

2

=

(

�

+

�

)

2

−

2

�

�

=

3

2

−

2

×

5

=

9

−

10

=

−

1.

(

�

�

)

�

3

+

�

3

=

(

�

+

�

)

3

−

3

�

�

(

�

+

�

)

=

3

3

−

3

(

5

)

(

3

)

=

27

−

45

=

−

18.

x

2

−3x+5=0

α+β=3

αβ=5

(i)α

2

+β

2

=(α+β)

2

−2αβ

=3

2

−2×5

=9−10

=−1.

(ii)α

3

+β

3

=(α+β)

3

−3αβ(α+β)

=3

3

−3(5)(3)

=27−45

=−18.