Question

If α and β are zeroes of 4x²+3x+7 , then find the value of (1/α³+1/β³).​

Answer 1

Answer:

On comparing with ax^2 + bx + c, we get a = 4, b = 3, c = 7. = > αβ = 7/4. = > -3/7. Therefore, the value of 1/α + 1/β = (-3/7).

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Answer 2

Solution :-

Given : f(x) = 3x² - 4x + 1

Sum of zeros = α + β = 4/3

Product of zeros = α β = 1/3

a) α/β + β/α

= [α² + β²]/βα

= [(α + β)² - 2αβ]/ αβ

= [(4/3)² - 2 × 1/3]/[1/3]

= [16/9 - 2/3]/[1/6]

= [(16 - 6)/9]/[1/3]

= (10 × 3)/(1 × 9)

= 10/3 ans.

b) α²/β + β²/α

= [α³ + β³]/α β

= [(α + β)³ - 3αβ(α + β)]/αβ

= [(4/3)³ - 3 × 1/3(4/3)]/1/3

= [64/27 - 4/3]/[1/3]

= [(64 - 36)/27 ]/[1/3]

= (28 × 3)/27

= 28/9 ans