If the zeroes of the quadratic polynomial x2 + (a + 1) x + b ar

1.	If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and –3, then(A) a = –7, b = –1 (B) a = 5, b = –1(C) a = 2, b = – 6 (D) a = 0, b = – 6
Answer» (D) a = 2, b = – 6 Explanation: Zeroes of a polynomial is all the values of x at which the polynomial is equal to zero. 2 and - 3 are the zeroes of the polynomial p(x) = x² + (a + 1)x + b i.e. p(2) = 0 and p(- 3) = 0 p(2) = (2)² + (a + 1)(2) + b = 0 = 4 + 2a + 2 + b = 0 = 6 + 2a + b = 0 (1) P(- 3) = (- 3)² + 9 + (a + 1)(- 3) + b = 0 = 9 - 3a - 3 + b = 0 = 6 - 3a + b = 0 (2) Equating (1) = (2), as both the equations are equal to zero. Hence both equations are equal to each other. 6 + 2a + b = 6 - 3a + b = 5a = 0 ⇒ a = 0 Putting the value of ‘a’ in (1) 6 + 2(0) + b = 0 ⇒ b = - 6 OR The equation of a quadratic polynomial is given by x² - (sum of the zeroes) x + (product of the zeroes) Sum of the zeroes = - (coefficient of x) ÷ coefficient of x² ⇒ sum of the zeroes = - (a + 1) = 2 - 3 = - a - 1 = - 1 + 1 = - a = - a = 0 ⇒ a = 0 Product of the zeroes = constant term ÷ coefficient of x² ⇒ b = product of the zeroes = 2(- 3) = 6

If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and –3, then(A) a = –7, b = –1 (B) a = 5, b = –1(C) a = 2, b = – 6 (D) a = 0, b = – 6

Answer»

(D) a = 2, b = – 6

Explanation:

Zeroes of a polynomial is all the values of x at which the polynomial is equal to zero.

2 and - 3 are the zeroes of the polynomial p(x) = x² + (a + 1)x + b

i.e. p(2) = 0 and p(- 3) = 0

p(2) = (2)² + (a + 1)(2) + b = 0

= 4 + 2a + 2 + b = 0

= 6 + 2a + b = 0 (1)

P(- 3) = (- 3)² + 9 + (a + 1)(- 3) + b = 0

= 9 - 3a - 3 + b = 0

= 6 - 3a + b = 0 (2)

Equating (1) = (2), as both the equations are equal to zero. Hence both equations are equal to each other.

6 + 2a + b = 6 - 3a + b

= 5a = 0

⇒ a = 0

Putting the value of ‘a’ in (1)

6 + 2(0) + b = 0

⇒ b = - 6

The equation of a quadratic polynomial is given by x² - (sum of the zeroes) x + (product of the zeroes)

Sum of the zeroes = - (coefficient of x) ÷ coefficient of x²

⇒ sum of the zeroes = - (a + 1)

= 2 - 3 = - a - 1

= - 1 + 1 = - a

= - a = 0

⇒ a = 0

Product of the zeroes = constant term ÷ coefficient of x²

⇒ b = product of the zeroes

= 2(- 3)

= 6