Question

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

Answer 1

Step-by-step explanation:

as we know

α+β = -b/a.

αβ =c/a.

here a = 1 b= a+1 c= b.

so,α+β = -b/a

=》-(a+1)/1 = 2+(-3)

-(a+1) = -1.

both LHS AND RHS MINUS WILL GET CANCELLED...

a+1 = 1.

a=1-1

a=0.

αβ =c/a.

(2)(-3)= b/1.

-6 = b

b= -6.

therefore a = 0 and b= -6

Answer 2

Answer:

Hello Dear User__________

Here is Your Answer...!!

____________________

Step-by-step explanation

$Given \ p(x)=x^{2}+(a+1)x+b\\\\Also \ given \ \alpha=2 \ and \ \beta=-3\\\\we \ know \ that \ \alpha+\beta=\frac{-b}{a}\\\\putting \ value \ here\\\\2+(-3)=\frac{-(a+1)}{1}\\\\-1=-(a+1)\\\\a=1-1\\\\a=0\\\\Now \ \alpha \beta=\frac{c}{a}\\\\ 2*-3=\frac{b}{1}\\\\b=-6$

Hope it is clear to you.