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)

___________________________