Question

α and β are the zeroes of the polynomial x2 – px + q, then find the value of (i) /+ () αβ3 + βα3 (iii) α3β2 + α2 β3​

Answer 1

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Answer 2

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Question:-

If α & β are the roots of the polynomial 3x² + 2x - 6 , then find the value of

i) α - β

ii) α² + β²

iii) α³ + β ³

iv) 1/α + 1/β

Answer:-

Given:

α & β are the roots of 3x² + 2x - 6.

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

Let,

a = 3

b = 2

c = - 6

We know that,

Sum of the roots = - b/a

⟹ α + β = - 2/3 -- equation (1)

Product of the roots = c/a

⟹ αβ = - 6/3 = - 2 -- equation (2)

We have to find:

i) α - β

We know that,

(a - b)² = (a + b)² - 4ab

So,

⟹ (α - β)² = (α + β)² - 4αβ

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

⟹ (α - β)² = (- 2/3)² - 4( - 2)

⟹ (α - β)² = 4/9 + 8

⟹ (α - β)² = (4 + 72)/9

⟹ (α - β)² = 76/9

⟹ (α - β) = √(76/9)

⟹ (α - β) = 2√19/3

Simple Method:-

If α , β are the roots of ax² + bx + c = 0 ;

(α - β) = √(b² - 4ac) / |a|

⟹ α - β = √2² - 4(3)( - 6) / |3|

⟹ α - β = √4 + 72 / 3

⟹ α - β = √76/3

⟹ α - β = 2√19/3

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ii) α² + β²

We know,

a² + b² = (a + b)² - 2ab

So,

⟹ α² + β² = (α + β)² - 2αβ

⟹ α² + β² = ( - 2/3)² - 2( - 2)

⟹ α² + β² = 4/9 + 4

⟹ α² + β² = (4 + 36)/9

⟹ α² + β² = 40/9

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iii) α³ + β³

We know,

a³ + b³ = (a + b)(a² + b² - ab)

⟹ α³ + β³ = (α + β)(α² + β² - αβ)

⟹ α³ + β³ = (- 2/3) [ 40/9 - (- 2) ]

⟹ α³ + β³ = (- 2/3) (40/9 + 2)

⟹ α³ + β³ = (- 2/3) (40 + 18) / 9

⟹ α³ + β³ = (- 2/3) (58/9)

⟹ α³ + β³ = - 116/27

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iv) 1/α + 1/β

Taking LCM we get,

⟹ (β² + α²) / αβ

⟹ (40/9) / ( - 2)

⟹ (40/9) * ( - 1/2)

⟹ 1/α + 1/β = - 20/9.