Question

if alpha and beta are zeros of 3x^2-x-4.Then find 1/alpha+1/beta and (1/alpha)^2+(1/beta)^2​

Answer 1

Answer:-

Given Polynomial => 3x² - x - 4.

Let a = 3 ; b = - 1 and c = - 4 and alpha, beta are taken as p & q.

We know that,

Sum of the zeroes = - b/a

→ p + q = -( - 1)/3

→ p + q = 1/3 -- equation (1)

Product of the zeroes = c/a

→ pq = - 4/3 -- equation (2).

Now , Taking LCM from the finding value,

→ (1/p) + (1/q) = (q + p)/pq

→ (p + q)/pq

Putting the values we get,

→ (1/3)/(- 4/3)

→ (1/3)(3/- 4)

→ 1/p + 1/q = - 1/4

We know that,

a² + b² = (a + b)² - 2ab

→ (1/p)² + (1/q)² = (1/p + 1/q)² - 2(1/p)(1/q)

→ (1/p)² + (1/q)² = (- 1/4)² - 2(1/pq)

→ (1/p)² + (1/q)² = 1/16 - 2(1/(- 4/3))

→ (1/p)² + (1/q)² = 1/16 + (3/2)

→ (1/p)² + (1/q)² = (1 + 24)/16

→ (1/p)² + (1/q)² = 25/16

Answer 2

Answer:

1/α+1/β=-1/4

1/α^2+1/β^2=1/16

Step-by-step explanation:

α+β=-b/a=-(-1/3)=1/3

αβ=c/a=(-4/3)

1. 1/α+1/β=α+β/αβ=(1/3)/(-4/3)=(-1/4)

2.1/α^2+β^2=α^2+β^2/(αβ)^2 =(α+β)^2-2αβ/(αβ)^2

=(1/3)^2-2*(-4/3)/(-4/3)^2

=(1/9+8/3)/(16/9)

=(25/9)/(16/9)

=25/16