Question

if alpha beta gamma are the zeros of polynomial ax³+bx²+cx+d. find the value of 1/alpha+1/beta+1/gamma.​

Answer 1

Answer: -c/d

We are given a cubic polynomial as,

• p(x) = ax³ + bx² + cx + d

Let α, β and λ be the zeroes of this polynomial, then we have to find the value of,

• 1/α + 1/β + 1/λ

Looking at the expression, we can say that we have to find the sum of the reciprocals of the zeroes but we can't find them as the coefficients are in variable. So, Let's simplify the expression:

⇒ 1/α + 1/β + 1/λ

⇒ (βλ + λα + αβ)/(αβλ)

⇒ (αβ + βλ + λα) / (αβλ) ...(i)

Now, The impossible looking expression is expressed in the form of Sum of products of two roots and product of the three roots. Now, we need to find these.

As we know, The sum of products of two roots is given by:

⇒ αβ + βλ + λα = (Coefficient of x) / (Coefficient of x³)

⇒ αβ + βλ + λα = c/a ...(ii)

Also, Product of all the three roots is given by:

⇒ αβλ = -(Constant term) / (Coefficient of x³)

⇒ αβλ = -d / a ...(iii)

Substituting the values in eq.(i) from eq.(ii) and eq.(iii) , we get

⇒ (c/a) / (- d/a)

⇒ c/a × - a/d

⇒ - c/d

Hence, The value of the expression 1/α + 1/β + 1/λ is -c/d.

Answer 2

ANSWER:

Given:

• α, β and γ are zeroes of polynomial, p(x) = ax³ + bx² + cx + d.

To Find:

• Value of 1/α + 1/β + 1/γ

Solution:

We are given a polynomial p(x) = ax³ + bx² + cx + d.

The zeroes of p(x) are α, β and γ.

So,

⇒ Sum of zeroes taken 2 at a time = (coefficient of x)/(coefficient of x³)

⇒ Sum of zeroes taken 2 at a time = (c)/(a) --------(1)

⇒ Product of zeroes = (constant)/(coefficient of x³)

⇒ Product of zeroes = (-d)/(a) --------(2)

Now,

⇒ 1/α + 1/β + 1/γ

Taking LCM,

⇒ 1/α + 1/β + 1/γ = (αβ + βγ + αγ)/(αβγ)

Here,

⇒ (αβ + βγ + αγ) = Sum of zeroes taken 2 at a time.

And,

⇒ (αβγ) = Product of zeroes.

From (1) & (2),

⇒ 1/α + 1/β + 1/γ = (c/a)/(-d/a)

⇒ 1/α + 1/β + 1/γ = (ca/-da)

⇒ 1/α + 1/β + 1/γ = -c/d

So, the value of 1/α + 1/β + 1/γ is -c/d.