Question

please help me to find out the answer​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

If α,β are zeroes of the polynomial x² - 5x + 6, form a quadratic polynomial whom zeroes are 2α and β + 1.

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

Let assume that

$\rm :\longmapsto\:p(x) = {x}^{2} - 5x + 6$

Let we factorize this polynomial using splitting of middle terms.

$\rm :\longmapsto\:p(x) = {x}^{2} - 3x - 2x + 6$

$\rm :\longmapsto\:p(x) = x(x - 3) - 2(x - 3)$

$\rm :\longmapsto\:p(x) = (x - 3)(x - 2)$

$\rm :\implies\:3, \: 2 \: are \: zeroes \: of \: p(x)$

But, it is given that

$\rm :\longmapsto\: \alpha, \: \beta \: are \: zeroes \: of \: p(x)$

So, two cases arises.

Case :- 1 When α = 3 and β = 2

and

Case :- 2 When α = 2 and β = 3

Now, Consider

Case - 1

$\rm :\longmapsto\: \alpha = 3 \: \: and \: \: \beta = 2$

Now, we have to find a quadratic polynomial whom zeroes are 2α and β + 1.

So, zeroes of the required polynomial are 2× 3 = 6 and 2 + 1 = 3.

So, Sum of zeroes, S = 6 + 3 = 9

and

Product of zeroes, P = 6 × 3 = 18

So, required Quadratic polynomial is given by

$\rm :\longmapsto\:f(x) = k\bigg[ {x}^{2} - Sx + P\bigg], \: where \: k \: \ne \: 0$

On substituting the values of S and P, we get

$\purple{\bf :\longmapsto\:f(x) = k\bigg[ {x}^{2} - 9x + 18\bigg], \: where \: k \: \ne \: 0}$

Case - 2

$\rm :\longmapsto\: \alpha = 2 \: \: and \: \: \beta = 3$

Now, we have to find a quadratic polynomial whom zeroes are 2α and β + 1.

So, zeroes of the required polynomial are 2× 2 = 4 and 3 + 1 = 4.

So, Sum of zeroes, S = 4 + 4 = 8

and

Product of zeroes, P = 4 × 4 = 16

So, required Quadratic polynomial is given by

$\rm :\longmapsto\:f(x) = k\bigg[ {x}^{2} - Sx + P\bigg], \: where \: k \: \ne \: 0$

On substituting the values of S and P, we get

$\purple{\bf :\longmapsto\:f(x) = k\bigg[ {x}^{2} - 8x + 16\bigg], \: where \: k \: \ne \: 0}$

Additional Information :-

1. For Cubic Polynomial

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

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

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

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

2. For a quadratic polynomial

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

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

$\boxed{ \bf{ \: \alpha \beta = \dfrac{c}{a}}}$