Question

If α,β,Y are the zeroes of the cubic polynomial,ax³+bx²+cx+d, then 1/α+1/β+1/Y​

Answer 1

To Find :-

• The value of 1/α + 1/β + 1/γ.

Solution :-

Given,

• α,β,γ are the zeroes of the cubic polynomial ax³+bx²+cx+d.

As we know that,

↪ α + β + γ = -b/a

↪ αβ + βγ + γα = c/a

↪ αβγ = -d/a

Now, we need to find the value of 1/α + 1/β + 1/γ.

↪ 1/α + 1/β + 1/γ

↪ αβ + βγ + γα / αβγ

[ °.° αβ + βγ + γα = c/a ]

[ °.° αβγ = -d/a ]

↪ (c/a) / (-d/a)

↪ c/a × a/-d

↪ c/-d

Therefore,

The value of 1/α + 1/β + 1/γ is c/-d.