Question

find the polynomial whose zeros are reciprocal of the zeros of the polynomial 2x square + 3 x minus 6

Answer 1

here is u r answer

hope u understand

Answer 2

Answer:

$x^2 - (2 +(\frac{4}{-3+\sqrt{57}}) + (\frac{4}{-3-\sqrt{57}}) )x + (1 +(\frac{4}{-3+\sqrt{57}}) )(1 + (\frac{4}{-3-\sqrt{57}})) = 0$

Step-by-step explanation:

Given : Polynomial $2x^2+3x-6$

To find : The polynomial whose zeros are reciprocal of the zeros of the given polynomial.

Solution : First we find the roots of the given polynomial.

Equate the given polynomial to zero to find the roots

$2x^2+3x-6=0$

Apply Discriminant and the solution is in the form

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

where a= 2 , b=3 , c=-6

$x= \frac{-3\pm\sqrt{3^2-4(2)(-6)}}{2(2)}$

$x= \frac{-3\pm\sqrt{9+48}}{4}$

$x= \frac{-3\pm\sqrt{57}}{4}$

Therefore, $x= \frac{-3+\sqrt{57}}{4}$ , $x= \frac{-3-\sqrt{57}}{4}$

Reciprocal of Roots are  $x= \frac{4}{-3+\sqrt{57}}$ , $x= \frac{4}{-3-\sqrt{57}}$

To form a polynomial with roots 1 + α and 1 + β. This yields the equation,

$\alpha= \frac{4}{-3+\sqrt{57}}$ , $\beta= \frac{4}{-3-\sqrt{57}}$

New equation = $(x-(1+\alpha))(x-(1+\beta )$

$x^2 - (2 +\alpha + \beta )x + (1 + \alpha )(1 + \beta ) = 0$

Put value of α and β

$x^2 - (2 +(\frac{4}{-3+\sqrt{57}}) + (\frac{4}{-3-\sqrt{57}}) )x + (1 +(\frac{4}{-3+\sqrt{57}}) )(1 + (\frac{4}{-3-\sqrt{57}})) = 0$

This is the new equation.