Question

If alpha and beta are zeros of polynomial x^2-x-2 find the equation where zeroes are 2 alpha+1 and 2 beta + 1

Answer 1

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

Given that

$\red{\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: {x}^{2} - x - 2}$

We know that,

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

$\rm :\longmapsto\: \alpha + \beta = - \: \dfrac{( - 1)}{1} = 1$

And

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

$\rm :\implies\: \alpha \beta = \dfrac{ - 2}{1} = - 2$

Now,

We have to find a quadratic polynomial whose zeroes are

$\rm :\longmapsto\:2 \alpha + 1 \: and \: 2 \beta + 1$

Let Consider,

Sum of the roots, S

$\rm \: = \: \: \:2 \alpha + 1 \: + \: 2 \beta + 1$

$\rm \: = \: \: 2( \alpha + \beta ) + 2$

$\rm \: = \: \: 2( 1 ) + 2$

$\rm \: = \: \: 2+ 2$

$\rm \: = \: \: 4$

Now, Consider Product of the roots, P

$\rm \: = \: \: (2 \alpha + 1)(2\beta + 1)$

$\rm \: = \: \: 4 \alpha \beta + 2 \alpha + 2 \beta + 1$

$\rm \: = \: \: 4 \alpha \beta + 2( \alpha + \beta) + 1$

$\rm \: = \: \: 4( - 2) + 2 + 1$

$\rm \: = \: \: - 8 + 3$

$\rm \: = \: \: - 5$

So, The required Quadratic polynomial is

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

So, on substituting the values of S and P, we get

${\bf :\longmapsto\:f(x) = k\bigg( {x}^{2} - 4x - 5 \bigg), \: where \: k \ne \: 0}$

Additional Information :-

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

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

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

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