Question

if alpha and beta are zeros of x2 +4x +3 , form polynomial whose zeros are 1+alpha/beta and 1+ beta /alpha

Answer 1

Hi Mate !!


Given equation :- x² + 4x + 3


Factorising it by middle term splitting :-

x² + 4x + 3

x² + 3x + x + 3

x ( x + 3 ) + 1 ( x + 3 )

( x + 3 ) ( x + 1 )


• ( x + 3 ) = 0

x = ( - 3 )


• ( x + 1 ) = 0

x = ( - 1 )


$so \: \: \alpha = ( -3) \: \: \: and \: \: \: \: \beta = ( - 1)$

• The zeros of new equation are :-


$\frac{1 + \alpha }{ \beta } \: \: \: and \: \: \: \frac{1 + \beta }{ \alpha }$


$\frac{1 + \alpha }{ \beta } = \frac{1 - 3}{ - 1} = \frac{ - 2}{ - 1} = 2$


$\frac{1 + \beta }{ \alpha } = \frac{1 - 1}{ - 3} = \frac{0}{ - 3} = 0$

So, the Zeros of new equation are 2 and 0

• Sum of the Zeros are :-

0 + 2 = 2

• Product of the Zeros are :-

0 × 2 = 0


♯ To form the quadratic equation we have formula as :-

x² - ( sum of Zeros )x + (product of Zeros)

Putting value in it !!

x² - 2x + 0

So, the required quadratic equation is
x² - 2x .

Answer 2

Answer:

The polynomial whose roots are $1+\frac{\alpha }{\beta }$  and  $1+\frac{\beta }{\alpha }$

= $\frac{1}{3}[3x^2 - 16x +16]$

Step-by-step explanation:

Given,

α and β are the roots of the polynomial $x^2 +4x +3$  = 0

To find,

The polynomial whose roots are $1+\frac{\alpha }{\beta }$ and $1+\frac{\beta }{\alpha }$

Recall the formula

If the roots of a quadratic polynomial are given, then the equation of the polynomial is given by

x² - ( sum of roots )x + Product of roots

Solution:

α and β are the roots of the polynomial x² + 4x +3 = 0

x² + 4x +3 = 0  ⇒ (x+3)(x+1) = 0

x = -3 and x = -1

Hence the roots of the polynomial are -3 and -1

Then we have α = -3 and β = -1

To find the polynomial whose roots are $1+\frac{\alpha }{\beta }$  and  $1+\frac{\beta }{\alpha }$

$1+\frac{\alpha }{\beta }$  = 1+ $\frac{-3}{-1}$ = 1+3 = 4

$1+\frac{\beta }{\alpha }$ = 1+ $\frac{-1}{-3}$ = $\frac{4}{3}$

Sum of roots = $1+\frac{\alpha }{\beta }$  +  $1+\frac{\beta }{\alpha }$  = 4+ $\frac{4}{3}$ = $\frac{16}{3}$

Product of roots = 4 ×  $\frac{4}{3}$ =  $\frac{16}{3}$

∴The required polynomial = x² - ( sum of roots )x + Product of roots

= x² - $\frac{16}{3}$x + $\frac{16}{3}$

= $\frac{1}{3}[3x^2 - 16x +16]$

∴ The polynomial whose roots are $1+\frac{\alpha }{\beta }$  and  $1+\frac{\beta }{\alpha }$

= $\frac{1}{3}[3x^2 - 16x +16]$

#SPJ2