Question

If alpha and beta are zeroes of poynomial fx= 6xsquare+x-1 the find value alpha/beta+bta/alpha+2(1/alpha+1bta

Answer 1

f ( x ) = 6x² + x - 1

By Middle Term Factorisation

f ( x ) = 6x² + 3x - 2x - 1

f ( x ) = 3x ( 2x + 1 ) - 1 ( 2x + 1 )

f ( x ) = ( 3x - 1 ) ( 2x + 1 )

To find the zeroes, f ( x ) = 0

→ 0 = ( 3x - 1 ) ( 2x + 1 )

Using Zero Product Rule

→ 3x - 1 = 0 and 2x + 1 = 0

→ x = 1/3 and x = - 1/2

Let the zeroes be α and β.

•°• α = 1/3, β = - 1/2

Now,

(α/β) + (β/α) + 2 [ (1/α) + (1/β) ]

Putting values, we get

→ $\sf{ {\dfrac{ {\dfrac{1}{3}} }{ {\dfrac{ - 1}{2}} }} + {\dfrac{ {\dfrac{ - 1}{2}} }{ {\dfrac{1}{3}} }} + 2 [{\dfrac{1}{3}} + {\dfrac{ - 1}{2}} ]}$

On solving this, we get

→ - 1/6

Hence, the answer is - 1/6.