Choose the correct answer from the given four options in the following questions:
If the zeroes of the quadratic polynomial $ x^{2}+(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 $

Given:

The zeroes of the quadratic polynomial \( x^{2}+(a+1) x+b \) are 2 and \( -3 \)

To do:

We have to find the values of $a$ and $b$.

Solution:

$x^2+(a+1)x+b$ is the quadratic polynomial.

$2$ and $-3$ are the zeros of the quadratic polynomial.

Thus,

Sum of the zeroes$=2+(-3)=-\frac{a+1}{1}$

$\Rightarrow -\frac{a+1}{1}=-1$

$\Rightarrow a+1=1$

$\Rightarrow a=0$

Also, Product of the zeroes$=2\times (-3)=\frac{b}{1}$

$\Rightarrow b=-6$

Thus, $a=0,\ b=-6$.

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Updated on: 10-Oct-2022

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