Question

if a. b and c are zeros of the polynomial x^3 + 3x^2 + 10x - 24 then find the value of 1/a + 1/b +1/c​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:a,b,c \: are \: zeroes \: of \: {x}^{3} + {3x}^{2} + 10x - 24$

Now, we know that

$\boxed{\red{\sf a + b + c=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

So,

$\bf\implies \:a + b + c= - \dfrac{3}{1} = - 3$

Also, we know that

$\boxed{\red{\sf ab + bc + ca=\frac{coefficient\ of\ {x}^{2} }{coefficient\ of\ x^{3}}}}$

So,

$\bf\implies \:ab+ bc + ca= \dfrac{10}{1} = 10$

Also, we know that

$\boxed{\red{\sf abc= - \: \frac{Constant}{coefficient\ of\ x^{3}}}}$

So,

$\bf\implies \:abc= - \dfrac{( - 24)}{1} = 24$

Now

Consider,

$\rm :\longmapsto\:\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$

$\rm \: = \: \: \dfrac{bc + ca + ab}{abc}$

$\rm \: = \: \: \dfrac{ab + bc + ca}{abc}$

$\rm \: = \: \: \dfrac{10}{24}$

$\rm \: = \: \: \dfrac{5}{12}$

Hence,

The value of

$\bf :\longmapsto\:\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{5}{12}$

Additional Information :-

1. Important identities :-

$\boxed{ \rm \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2\alpha \beta }$

$\boxed{ \rm \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3\alpha \beta( \alpha + \beta ) }$

$\boxed{ \rm \: {( \alpha - \beta )}^{2} = {( \alpha + \beta )}^{2} - 4 \alpha \beta }$

2.

$\rm :\longmapsto\: \alpha , \: \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c, \: then$

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

and

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$