Question

if alpha, beta and gamma are the roots of
the equation x^3 - 4x^2+6X-4 = 0.
the value of 1/alpha + 1/beta + 1/gamma is​

Answer 1

Topic

• Polynomials- Vieta's Formula

It refers to the method of finding the values of expression for zeroes.

Solution

① Finding the relation between roots.

Let $P(x)=x^{3}-4x^{2}+6x-4$. Given zeroes are $\alpha ,\beta ,\gamma$, so let the new zeroes be $\alpha ',\beta ',\gamma '$.

The relation between two groups of zeroes is the following.

➊ $\alpha '=\dfrac{1}{\alpha }$

➋ $\beta '=\dfrac{1}{\beta }$

➌ $\gamma '=\dfrac{1}{\gamma }$

② Finding the required answer.

The needed polynomial is $(x-\alpha ')(x-\beta ')(x-\gamma ')$.

We can find the polynomial having new zeros with the least degree, and such polynomial is $x^{3}P\left(\dfrac{1}{x} \right)$.

$\implies x^{3}P\left(\dfrac{1}{x} \right)=-4x^{3}+6x^{2}-4x+1$ $^{\bold{[1]}}$

Hence, the sum of the multiplicative inverses is the following.

$\implies \dfrac{1}{\alpha } +\dfrac{1}{\beta } +\dfrac{1}{\gamma } =-\dfrac{6}{-4} =\boxed{\dfrac{3}{2} }$

More Information

$^{\bold{[1]}}$ If you see this carefully, you'll find that the coefficients are in ascending order. This happened because we substituted $x$ with $\dfrac{1}{x}$. This happens when we find a new polynomial with inverse roots.

For example, all the roots of $3x^{2}+5x+9=0$ are multiplicative inverses of $9x^{2}+5x+3=0$. Try this once.

Answer 2

Given :-

x³ - 4x² + 6x - 4 = 0

To Find :-

Value of 1/α + 1/β + 1/γ

Solution :-

We know that

α + β + γ = -b/a

α + β + γ = -(-4)/1

α + β + γ = 4/1

α + β + γ = 4

αβγ = -d/a

αβγ =  -(-4)/1

αβγ = 4

αβ + βγ + αγ = c/a

αβ + βγ + αγ = 6/1

αβ + βγ + αγ = 6

Finding value of 1/α + 1/β + 1/γ

1/α + 1/β + 1/γ

αβ + βγ + αγ/αβγ

6/4

3/2