Question

If the zeroes of the quadratic polynomial x? +(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

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

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

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

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

Answer 2

Answer:

⠀⠀⠀⌬ Polynomial = x² + (a + 1)x + b

⠀⠀⠀⌬ Zeroes = 2 and - 3

⠀⠀⠀⌬ A = 1 , B = (a + 1) , C = b

$\underline{\bigstar\:\textsf{According to the given Question :}}$

Sum of Zeroes

$:\implies\sf \alpha+\beta=\dfrac{-B}{A}\\ \\ \\ :\implies\sf\ -3+2=\dfrac{-(a+1)}{1}\\ \\ \\ :\implies\sf\ -1= -(a+1)\\\\\\:\implies\sf 1 = a + 1\\ \\ \\ :\implies\sf\ 1-1=a\\ \\ \\ :\implies\sf\ a=0$

⠀

Product of Zeroes

$:\implies\sf \ \alpha\beta=\dfrac{C}{A}\\ \\ \\ :\implies\sf\ 3\times -2= \dfrac{b}{1}\\ \\ \\ :\implies\sf\ b= - \:6$

⠀

$\therefore\:\underline{\textsf{Hence, required answer is d) \textbf{a = 0 and b = - 6}}}.$