Question

If −2 and 3 are the zeros of the quadratic polynomial x² + (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

Answer 1

$$\begin{gathered} \left(\text{c}\right)\mathit{\text{a=-2, b=}}\text{-6} \\ \text{Given:}-2\text{and 3 are the zeroes of}{x}^2+(a+1)x+b. \\ \begin{array}{l}{\mathrm{Now},(-2)}^2+(a+1)\times (-2)+b=0=>4-2a-2+b=0\\ \text{=>}b-2a=-2...\left(1\right)\end{array} \\ \mathrm{Also},3^2+(a+1)\times 3+b=0=>9+3a+3+b=0 \\ \text{=>}b+3a=-12...\left(2\right) \\ \text{On subtracting}\left(1\right)\text{from}\left(2\right),\text{we get}a=-2 \\ \text{∴}b=-2-4=-6[\mathrm{From}(1\left)\right] \end{gathered}$$