Question

please tell me the answer 18,19​

Answer 1

-: Question 18 :-

Given :-

α , β and γ are the zeroes of the polynomial f ( x ) = ax³ + bx² + cx + d

To Find :-

The Value of α² + β² + γ²

Used Concepts :-

We knows that , for a cubic polynomial ax³ + bx² + cx + d with zeroes " α , β and γ " :-

• α + β + γ = -b/a

• α β γ = -d/a

• αβ + βγ + γα = c/a

Solution :-

Now , For f ( x ) = ax³ + bx² + cx + d

α + β + γ = -b/a

α β γ = -d/a

αβ + βγ + γα = c/a

Now , We have to find the value of α² + β² + γ². We will use a algebraic identity to find it .

=> ( α + β + γ )² = α² + β² + γ² + 2αβ + 2βγ + 2γα

=> ( α + β + γ )² = ( α² + β² + γ² ) + 2( αβ + βγ + γα )

=> Putting all the values we get,.

=> ( -b/a )² = ( α² + β² + γ² ) + 2( c/a )

=> b²/a² = ( α² + β² + γ² ) + 2c/a

=> ( α² + β² + γ² ) = b²/a² - 2c/a

=> ( α² + β² + γ² ) = b² - 2ac/a²

Henceforth , Our Required answer is ( d )

-: Question 19 :-

Given :-

α , β and γ are the zeroes of the polynomial f ( x ) = x³ - px² + qx - r

To Find :-

The Value of 1/αβ + 1/βγ + 1/γα

Solution :-

On comparing f ( x ) with the standard form of a cubic polynomial i.e ax³ + bx² + cx + d we get ,

a = 1 , b = - p , c = q and d = - r

Now , As α , β and γ are zeroes of f ( x ) . So ;

=> α β γ = -d/a = - ( - r )/1 = r

=> α + β + γ = -b/a = - ( - p )/1 = p

Now , We have to find value of 1/αβ + 1/βγ + 1/γα which can be simplified as follows :-

=> 1/αβ + 1/βγ + 1/γα

=> Taking L.C.M ;

=> γ + α + β/α × β × γ

=> α + β + γ/α × β × γ

=> Putting all values we get ;

=> p/r

Henceforth , Our Required Answer is ( b ) .

Additional Information :-

For a Bi-quadratic polynomial " ax⁴ + bx³ + cx² + dx + e " with zeroes " α , β , γ and δ " :-

• α + β + γ + δ = -b/a
• αβγδ = e/a
• ( α + β ) ( γ + δ ) + αβ + γδ = c/a
• αβ ( γ + δ ) + γδ ( α + β ) = -d/a

Answer 2

$\large\underline{\sf{Solution-18}}$

Given that,

$\rm :\longmapsto\: \alpha, \beta , \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d$

We know,

$\boxed{\red{\sf \alpha + \beta + \gamma =\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

$\bf\implies \: \alpha + \beta + \gamma = - \dfrac{b}{a}$

Also,

$\boxed{\red{\sf \alpha \beta + \beta \gamma + \gamma \alpha =\frac{-coefficient\ of\ {x}^{2} }{coefficient\ of\ x^{3}}}}$

$\bf\implies \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}$

Also,

$\boxed{\red{\sf \alpha \beta \gamma =\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha \beta \gamma = - \dfrac{d}{a}$

Consider,

$\rm :\longmapsto\: { \alpha }^{2} + { \beta }^{2} + { \gamma }^{2}$

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

$\rm \: = \: {\bigg(\dfrac{ - b}{a} \bigg)}^{2} - 2\bigg(\dfrac{c}{a} \bigg)$

$\rm \: = \: \dfrac{ {b}^{2} }{ {a}^{2} } - \dfrac{2c}{a}$

$\rm \: = \: \dfrac{ {b}^{2} - 2ac}{ {a}^{2} }$

Hence, Option (d) is correct.

$\large\underline{\bf{Solution-19}}$

Given that,

$\rm :\longmapsto\: \alpha, \beta , \gamma \: are \: zeroes \: of \: {x}^{3} - {px}^{2} + qx - r$

We know,

$\boxed{\blue{\sf \alpha + \beta + \gamma =\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

$\bf\implies \: \alpha + \beta + \gamma = - \dfrac{ ( - p)}{1} = p$

Also,

$\boxed{\blue{\sf \alpha \beta + \beta \gamma + \gamma \alpha =\frac{-coefficient\ of\ {x}^{2} }{coefficient\ of\ x^{3}}}}$

$\bf\implies \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{q}{1} = q$

Also,

$\boxed{\blue{\sf \alpha \beta \gamma =\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha \beta \gamma = - \dfrac{( - r)}{1} = r$

Consider,

$\rm :\longmapsto\:\dfrac{1}{ \alpha \beta } + \dfrac{1}{ \beta \gamma } + \dfrac{1}{ \gamma \alpha }$

$\rm \: = \: \dfrac{ \gamma + \beta + \alpha }{ \alpha \beta \gamma }$

$\rm \: = \: \dfrac{p}{r}$

Hence, Option (b) is correct.