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If $-2$ and 3 are the zeroes of the quadratic polynomial $x^{2}+(a+1) x+b$, then (a) $a=-2, b=6$ (b) $a=2, b=-6$ (c) $a=-2, b=-6$ (d) $a=2, b=6$
If $-2$ and 3 are the zeroes of the quadratic polynomial $x^{2}+(a+1) x+b$, then (a) $a=-2, b=6$ (b) $a=2, b=-6$ (c) $a=-2, b=-6$ (d) $a=2, b=6$
CBSE | Class 10 | Excercise 2D | Maths | Polynomials | RS Aggarwal
If $-2$ and 3 are the zeroes of the quadratic polynomial $x^{2}+(a+1) x+b$, then (a) $a=-2, b=6$ (b) $a=2, b=-6$ (c) $a=-2, b=-6$ (d) $a=2, b=6$

The correct option is option (c) $a=-2, b=-6$

$-2$ and 3 are the zeroes of $x^{2}+(a+1) x+b$.

$(-2)^{2}+(a+1) \times(-2)+b=0 \Rightarrow 4-2 a-2+b=0$

$\Rightarrow b-2 a=-2$ ….(1)

$3^{2}+(a+1) \times 3+b=0 \Rightarrow 9+3 a+3+b=0$

$\Rightarrow b+3 a=-12$ $\ldots(2)$

On subtracting (1) from (2), we get $\mathrm{a}=-2$

$\therefore b=-2-4=-6 \quad$ [From (1) $]$

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