Question:
If $-2$ and 3 are the zeros 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
Solution:
(c) $a=-2, b=-6$
Given: $-2$ and 3 are the zeroes of $x^{2}+(a+1) x+b$.
Now, $(-2)^{2}+(a+1) \times(-2)+b=0=>4-2 a-2+b=0$
$=>b-2 a=-2 \quad \ldots(1)$
Also, $3^{2}+(a+1) \times 3+b=0=>9+3 a+3+b=0$
$=>b+3 a=-12 \quad \ldots(2)$
On subtracting (1) from (2), we get $a=-2$
$\therefore b=-2-4=-6 \quad[\operatorname{From}(1)]$
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