If −2 and 3 are the zeros of the quadratic polynomial
Question:

If $-2$ and 3 are the zeros of the quadratic polynomial $x^{2}+(a+1) x+b$, then

(a) = −2, b = 6
(b) = 2, b = −6
(c) = −2, b = −6
(d) = 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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