Question

alpha and beta are the zeroes of the polynomial x^2-px-q then find the value of each of the following...
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Answer 1

Question:-

If α , β are the zeroes of the polynomial x² - px + q then find the value of each of the following.

i) α² + β²

ii) (α/β) + (β/α)

iii) α³ + β³

iv) α³β² + α²β³

v) αβ³ + α³β

vi) α - β

vii) α³ - β³

Answer:-

Given:

α , β are the zeroes of the polynomial x² - px + q

On comparing it with standard form of a quadratic equation i.e., ax² + bx + c = 0 ;

Let,

• a = 1
• b = - p
• c = q.

We know that,

Sum of the roots = - b/a

So,

⟹ α + β = - ( - p)/1

⟹ α + β = p -- equation (1)

Product of the roots = c/a

⟹ αβ = q -- equation (2)

We have to find:-

i) α² + β²

We know that,

a² + b² = (a + b)² - 2ab

So,

⟹ α² + β² = (α + β)² - 2αβ

Putting the respective values from equations (1) & (2) we get,

⟹ α² + β² = (p)² - 2(q)

⟹ α² + β² = p² - 2q

___________________________

ii) (α/β) + (β/α)

Taking LCM we get,

⟹ (α² + β²) / αβ

Putting the respective values we get,

⟹ (α/β) + (β/α) = (p² - 2q)/q

___________________________

iii) α³ + β³

We know,

a³ + b³ = (a + b)³ - 3ab(a + b)

So,

⟹ α³ + β³ = (α + β)³ - 3αβ(α + β)

Putting the values we get,

⟹ α³ + β³ = (p)³ - 3(q)(p)

⟹ α³ + β³ = p³ - 3pq

___________________________

iv) α³β² + α²β³

Taking α²β² common we get,

⟹ α²β² (α + β)

⟹ (αβ)² (α + β)

⟹ (q)² (p)

⟹ α³β² + α²β³ = pq²

___________________________

v) αβ³ + α³β

Taking αβ common we get,

⟹ (αβ) (α² + β²)

Putting the respective values we get,

⟹ (q) (p² - 2q)

⟹ αβ³ + α³β = p²q - 2q²

___________________________

vi) α - β

We know that,

(a - b)² = a² + b² - 2ab

So,

⟹ (α - β)² = (α² + β²) - 2αβ

Putting the respective values we get,

⟹ (α - β)² = p² - 2q - 2q

⟹ (α - β)² = p² - 4q

⟹ α - β = √(p² - 4q)

___________________________

vii) α³ - β³

We know that,

a³ - b³ = (a - b)³ + 3ab(a - b)

So,

⟹ α³ - β³ = (α - β)³ + 3αβ(α - β)

⟹ α³ - β³ = (√p² - 4q)³ + 3(q)(√p² - 4q)

⟹ α³ - β³ = (p² - 4q)(√p² - 4q) + 3q (√p² - 4q)

Taking √p² - 4q common in RHS we get,

⟹ α³ - β³ = (√p² - 4q) (p² - 4q + 3q)

⟹ α³ - β³ = (√p² - 4q)(p² - q)

___________________________

Answer 2

Answer:

Given Polynomial : x² - px + q

• Here a = 1 , b = -p, c = q

Sum of zeroes:

$\alpha + \beta = \frac{ - b}{a} = \frac{ p}{1} = p$

Product of zeroes:

$\alpha \beta = \frac{c}{a} = \frac{q}{1} = q$

(I) α² + β² :-

${ \alpha}^{2} + {\beta}^{2} = {(\alpha + \beta)}^{2} - 2 \alpha \beta \\ \\ { \alpha}^{2} + {\beta}^{2} = {p}^{2} - 2q$

(II) α/β + β/α :-

$\frac{ \alpha}{ \beta} + \frac{ \beta}{ \alpha} = \frac{ { \alpha}^{2} + { \beta}^{2} }{ \alpha \beta} \\ \\ \frac{ \alpha}{ \beta} + \frac{ \beta}{ \alpha} = \frac{ {p}^{2} - 2q}{q}$

(III) α³ + β³:-

${ \alpha}^{3} + { \beta}^{3} = {( \alpha + \beta)}^{3} - 3 \alpha \beta( \alpha \: + \beta) \\ \\ { \alpha}^{3} + { \beta}^{3} = {p}^{3} - 3pq$

(IV) α³β² + α²β³:-

${ \alpha}^{3} { \beta}^{2} + { \alpha}^{2} { \beta}^{3} = { \alpha}^{2} { \beta}^{2} ( \alpha + \beta) \\ \\ { (\alpha \beta)}^{2} ( \alpha + \beta) = p {q}^{2}$

(V) αβ³ + α³β :-

$\alpha { \beta}^{3} + { \alpha}^{3} \beta = \alpha \beta( { \alpha}^{2} + {\beta}^{2} ) \\ \\ \alpha { \beta}^{3} + { \alpha}^{3} \beta = {p}^{2} q - 2 {q}^{2}$

(VI) α - β:-

${(\alpha - \beta) }^{2} = { \alpha}^{2} + { \beta}^{2} - 2 \alpha \beta \\ \\ {(\alpha - \beta ) }^{2} = {p}^{2} - 4q \\ \\ \alpha - \beta = \sqrt{ p²- 4q }$

(VII) α³ - β³ :-

${ \alpha}^{3} - { \beta}^{3} = {( \alpha - \beta)}^{3} + 3 \alpha \beta( \alpha - \beta) \\ \\ { \alpha}^{3} - { \beta}^{3} = \sqrt{( {p}^{2} - 4q)}( p² - q)$

Note:

• (VII) sum solution in attachment.