Question

If alpha beta gamma are the zeros of polynomial x3-px2+qx-r then find 1\alpha beta + 1\beta gamma + 1\alpha gamma 9


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Answer 1

Answer:

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Answer 2

Answer:

$\frac{1}{ \alpha \beta } + \frac{1}{ \beta \gamma } + \frac{1}{ \alpha \gamma } = \frac{p}{r}$

If

$\alpha ,\beta ,\gamma \: are \: the \: roots \: of \: cubic \: equation \: {x}^{3} - p {x}^{2} + qx - r \: = 0$

Step-by-step explanation:

1)Step 1:

Given α, β, γ are the roots of cubic equation x³- px²+qx-r = 0

Here as we can see

The coefficient of x³ = 1

The coefficient of x² = -p

The coefficient of x = q

The constant term = -r

2) Step 2:

The sum of roots is given by

$\alpha + \beta + \gamma = \frac{ - ( - p)}{1} = p$

3) Step 3:

The product of roots is given by

$\alpha \beta \gamma = \frac{ - ( - r)}{1} = r$

4) Step 4:

We have to find the value of

$\frac{1}{ \alpha \beta } + \frac{1}{ \beta \gamma } + \frac{1}{ \alpha \gamma }$

5) Step 5:

Simplifying it by taking L.C.M.

$= \frac{ \gamma + \alpha + \beta }{ \alpha \beta \gamma }$

Using the values obtained for sum and product of roots, we get

$= \frac{p}{r}$

Thus the value comes out to be p/r

$\frac{1}{ \alpha \beta } + \frac{1}{ \beta \gamma } + \frac{1}{ \alpha \gamma } = \frac{p}{r}$