If the zeroes of the quadratic polynomial x2 + (+ 1) are 2 and –3, then
If the zeroes of the quadratic polynomial x2 + (+ 1) are 2 and –3, then

(A) = –7, = –1 (B) = 5, = –1 (C) = 2, = – 6 (D) = 0, = – 6

(D) a = 2, b = – 6

Clarification:

As per the inquiry,

    \[x{}^\text{2}\text{ }+\text{ }\left( a+1 \right)x\text{ }+\text{ }b\]

Considering that, the zeroes of the polynomial = 2 and – 3,

At the point when x = 2

    \[2{}^\text{2}\text{ }+\text{ }\left( a+1 \right)\left( 2 \right)\text{ }+\text{ }b\text{ }=\text{ }0\]

    \[4\text{ }+\text{ }2a+2\text{ }+\text{ }b\text{ }=\text{ }0\]

    \[6\text{ }+\text{ }2a+b\text{ }=\text{ }0\]

    \[2a+b\text{ }=\text{ }-\text{ }6\text{ }\text{ }\text{ }\left( 1 \right)\]

At the point when x = – 3,

    \[\left( -\text{ }3 \right){}^\text{2}\text{ }+\text{ }\left( a+1 \right)\left( -\text{ }3 \right)\text{ }+\text{ }b\text{ }=\text{ }0\]

    \[9\text{ }\text{ }3a-3\text{ }+\text{ }b\text{ }=\text{ }0\]

    \[6\text{ }\text{ }3a+b\text{ }=\text{ }0\]

    \[-\text{ }3a+b\text{ }=\text{ }-\text{ }6\text{ }\text{ }\text{ }\left( 2 \right)\]

Deducting condition (2) from (1)

    \[2a+b\text{ }\text{ }\left( -\text{ }3a+b \right)\text{ }=\text{ }-\text{ }6-\left( -\text{ }6 \right)\]

    \[2a+b+3a-b\text{ }=\text{ }-\text{ }6+6\]

    \[5a\text{ }=\text{ }0\]

    \[a\text{ }=\text{ }0\]

Subbing the worth of ‘a’ in condition (1), we get,

    \[2a\text{ }+\text{ }b\text{ }=\text{ }-\text{ }6\]

    \[2\left( 0 \right)\text{ }+b\text{ }=\text{ }-\text{ }6\]

b = – 6

Subsequently, alternative (D) is the right reply.