Question

if alpha and beta are zeroes of polynomial x2-2x+3 ,find a polynomial whose roots are alpha minus 1,upon alpha plus 1 and beta minus 1 upon beta plus 1

Answer 1

hope it helps you out

Answer 2

Answer:

Required polynomial is k ( 3x² - 2x + 1 )

Step-by-step explanation:

Given: Quadratic polynomial,

x² - 2x + 3

let α and β are zeroes of polynomial.

Consider,

x² - 2x + 3 = 0

to find zeroes we use quadratic formula,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-2)\pm\sqrt{(-2)^2-4\times1\times3}}{2\times1}$

$x=\frac{2\pm\sqrt{4-12}}{2}$

$x=\frac{2\pm\sqrt{-8}}{2}$

$x=\frac{2\pm\sqrt{8}\times\sqrt{-1}}{2}$

$x=\frac{2\pm2\sqrt{2}i}{2}$

$x=1\pm\sqrt{2}i$

So, $\alpha=1+\sqrt{2}$   and [tex\beta=1-\sqrt{2}i[/tex]

New roots are

$\frac{\alpha-1}{\alpha+1}=\frac{1+\sqrt{2}i-1}{1+\sqrt{2}i+1}=\frac{\sqrt{2}i}{2+\sqrt{2}i}=\frac{\sqrt{2}i}{2+\sqrt{2}i}\times\frac{2-\sqrt{2}i}{2-\sqrt{2}i}=\frac{2\sqrt{2}i-2\times(-1)}{4-2\times(-1)}=\frac{1+\sqrt{2}i}{3}$

$\frac{\beta-1}{\beta+1}=\frac{1-\sqrt{2}i-1}{1-\sqrt{2}i+1}=\frac{-\sqrt{2}i}{2-\sqrt{2}i}=\frac{-\sqrt{2}i}{2-\sqrt{2}i}\times\frac{2+\sqrt{2}i}{2+\sqrt{2}i}=\frac{-2\sqrt{2}i-2\times(-1)}{4-2\times(-1)}=\frac{1-\sqrt{2}i}{3}$

Now Sum of new zeroes = $\frac{1+\sqrt{2}i}{3}+\frac{1-\sqrt{2}i}{3}$

                                         = $\frac{2}{3}$

Product of new zeroes =  $\frac{1+\sqrt{2}i}{3}\times\frac{1-\sqrt{2}i}{3}$

                                         = $\frac{1^2-(\sqrt{2}i)^2}{9}$

                                         = $\frac{1-2\times(-1)}{9}$

                                         = $\frac{1}{3}$

Thus, New polynomial is

$k(x^2-\frac{2}{3}x+\frac{1}{3})$

$\implies\:\:k(3x^2-2x+1)$

Therefore, Required polynomial is k ( 3x² - 2x + 1 )