If - 2 and 3 are the zeroes of the quadratic polynomial x2 + (a + 1)x

1.	If - 2 and 3 are the zeroes 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» (c) a = –2, b = –6 Given: –2 and 3 are the zeroes of x² + (a + 1) x + b. Now, (–2)² + (a + 1) × (–2) + b = 0 ⇒ 4 – 2a – 2 + b = 0 ⇒ b – 2a = –2 ….(1) Also, 3² + (a + 1) × 3 + b = 0 ⇒ 9 + 3a + 3 + b = 0 ⇒ b + 3a = –12 ….(2) On subtracting (1) from (2), we get a = –2 ∴ b = –2 – 4 = –6 [From (1)]

If - 2 and 3 are the zeroes 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»

(c) a = –2, b = –6

Given:

–2 and 3 are the zeroes of x² + (a + 1) x + b.

Now,

(–2)² + (a + 1) × (–2) + b = 0

⇒ 4 – 2a – 2 + b = 0

⇒ b – 2a = –2 ….(1)

Also, 3² + (a + 1) × 3 + b = 0

⇒ 9 + 3a + 3 + b = 0

⇒ b + 3a = –12 ….(2)

On subtracting (1) from (2),

we get a = –2

∴ b = –2 – 4 = –6 [From (1)]