Question

14. If a, B and y are the zeroes of the polynomial f(x) = x3 - px2 + x-r, then​

Answer 1

Value of 1/αβ + 1/βγ + 1/αγ = p/r.

Step-by-step explanation:

★ Given that :

• α,β and γ are the zeroes of the cubic Polynomial x³ - px² + x - r = 0

★ To find :

• Value of 1/αβ + 1/βγ + 1/αγ

★ Let :

◼ The general form of cubic polynomial is ax³ + bx² + cx + d = 0.

◼ Consider the given polynomial.

• a = 1
• b = - p
• c = 1
• d = - 1

★ Now :

$\bf\red{\rightarrow Sum\:of\:the\:zeroes : \alpha +\beta+\gamma = \cfrac{-b}{a} }$

$\sf \implies \alpha+\beta+\gamma = \cfrac{-(-p)}{1}$

$\sf \implies \alpha+\beta+\gamma = p$

$\bf\red{\rightarrow Sum\:of\:the\:product\:of\:the\:zeroes : \alpha \beta + \beta\gamma + \alpha\gamma= \cfrac{c}{a} }$

$\sf \implies \alpha\beta + \beta\gamma+\alpha\gamma = \cfrac{1}{1}$

$\sf \implies \alpha\beta + \beta\gamma+\alpha\gamma = 1$

$\bf\red{\rightarrow Product\:of\:the\:zeroes : \alpha \beta \gamma = \cfrac{c}{a} }$

$\sf \implies \alpha\beta\gamma = \cfrac{-(-r)}{1}$

$\sf \implies \alpha\beta\gamma = r$

★ Now :

Find out the value of :-

$\sf \cfrac{1}{\alpha\beta} + \cfrac{1}{\beta\gamma} + \cfrac{1}{\alpha\gamma}$

$\sf \implies \cfrac{1}{\alpha\beta} + \cfrac{1}{\beta\gamma} + \cfrac{1}{\alpha\gamma}$

$\sf \implies \cfrac{\alpha+\beta+\gamma}{\alpha\beta\gamma}$

• Substitute the zeroes.

$\sf \implies \cfrac{p}{r}$

$\underline{\boxed{\rm{\purple{\therefore Hence,\:the\:value\:of \: \: \:\cfrac{\alpha +\beta+\gamma}{\alpha\beta\gamma} = \cfrac{p}{r}.}}}}\:\orange{\bigstar}$