Question

If the zeros of the quadratic polynomial x² + (a+1)x + b are 2 and -3, then

(A) a = -7, b= -1
(B) a = 5, b = -1
(C) a = 2, b = -6
(D) a = 0, b = -6

Answer 1

Answer:

a = 0 and b = -6

Step-by-step explanation:

(x-2)(x+3) are the roots

On multiplying,

> x^2 + 3x - 2x - 6

> x^2 + x - 6

On comparing with x^2 + (a + 1)x + b

We get a = 0 and b = -6

Answer 2

$\mathsf{\underline{\red{Solution:- }}}$

we have

$\sf\ \alpha= 2\ \ \ and\ \ \beta= -3$

we have to find the value of a and b

As we know that :-

$\sf\ \alpha+\beta=\dfrac{-b}{a}\\ \\ \\ \sf\ \ \alpha\beta= \dfrac{c}{a}$

In the given polynomial

$\sf\ x^2+(a+1)x+b=0\\ \\ \\ \sf\ \ a= 1\ \ \ ;\ b= \ (a+1)\ \ ;\ \ c=b$

Now ,

$\sf\alpha+\beta=\dfrac{-b}{a}\\ \\ \\ :\implies\sf\ -3+2=\dfrac{-(a+1)}{1}\\ \\ \\ :\implies\sf\ \cancel-1= \cancel-(a+1)\\ \\ \\ :\implies\sf\ 1-1=a\\ \\ \\ :\implies\underline{\boxed{\red{\sf\ a=0}}}$

Again for b

$\sf\ \ \alpha\beta=\dfrac{c}{a}\\ \\ \\ :\implies\sf\ 3\times -2= \dfrac{b}{1}\\ \\ \\ :\implies\underline{\boxed{\purple{\sf\ b= -6}}}$

$\underline{\sf{\maltese\ Option (d)\ is\ the\ correct\ \ }}$