Math, asked by Roseviya, 3 months ago

If 3 and (-2) are the zeroes of the polynomial x²+(a+1)x-b, find the value of a and b​

Answers

Answered by tennetiraj86
2

Step-by-step explanation:

Given:-

3 and (-2) are the zeroes of the polynomial x^2+(a+1)x-b.

To find:-

Find the values of a and b ?

Solution:-

Given Polynomial = x^2+(a+1)x-b.

On Comparing this with the standard quadratic Polynomial ax^2+bx+c

a = 1

b= (a+1)

c = -b

Given zeroes are 3 and -2

We know that

Sum of the zeroes (α+β)= -b/a

=> 3+(-2) = - (a+1)/1

=> 3-2 = -(a+1)

=> 1= -a-1

=> -a-1 = 1

=> -a = 1+1

=> -a = 2

=> a = -2

The value of a = -2

and

Product of the zeroes (αβ) = c/a

=> 3×(-2) = -b/1

=> -6 = -b

=> -b = -6

=> b = 6

The Value of b = 6

Answer:-

The value of a = -2

The Value of b = 6

Used formulae:-

  • The standard quadratic Polynomial ax^2+bx+c

  • Sum of the zeroes (α+β)= -b/a

  • Product of the zeroes (αβ) = c/a
Similar questions