Question

if 3 and (-2) are the zeroes of the polynomial x^2+(a+1)x-b find the value of a and b​

Answer 1

Answer✔

x2+(a+1)x+b

A = 1; B = (a+1) C = b

Let alpha be 2 and beta be -3

alpha + beta = -1

-B/A = -1

-(a+1) = -1

-a - 1 = -1

a = 0

⭐Hope it helps⭐

Answer 2

Given:

• Quadratic Polynomial → x² + (a + 1)x - b
• Zeroes = 3, -2

To find:

Value of a and b = ?

Solution:

First, let's find out the sum of zeroes.

$\sf{Sum \: of \: zeroes = \dfrac{-b}{a}}$

$\to \: \sf{3-2=\dfrac{-(a+1)}{1}}$

$\to \: \sf{-a-1=1}$

$\to \: \sf{-a=1+1}$

$\to \: \sf{-a=2}$

$\therefore \boxed{\bf{a=-2}}$

Now for finding out the Product of zeroes;

$\sf{Product\: of \: zeroes = \dfrac{c}{a}}$

$\to \: \sf{3(-2)=\dfrac{-b}{1}}$

$\to \: \sf{-6=-b}$

$\therefore \boxed{\bf{b=6}}$