Question

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

Answer 1

Answer: a = 0, b = -6

Step-by-step explanation:

p(x) = x² + (a + 1)x + b

Zeroes of this polynomial are 2 and (-3). This means if we put 2 and (-3) as x in the above equation, we should get 0.

p(2) = 2² + (a + 1)2 + b = 0

4 + 2a + 2 + b = 0

2a + b + 6 = 0 (First Equation)

p(-3) = (-3)² + (a + 1)(-3) + b = 0

9 - 3a - 3 + b = 0

-3a + b + 6 = 0 (Second Equation)

Now, we have two equations and two variables. So we can find the values of a and b.

First Equation: 2a + b + 6 = 0

b = -2a - 6

Put the value of b obtained here in the second equation.

Second Equation: -3a + b + 6 = 0

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

-3a - 2a - 6 + 6 = 0

-5a = 0

a = 0

Now, find the value of b.

b = -2a - 6

b = -2(0) - 6

b = 0 - 6

b = -6

Answer 2

The value of a = 0, b = -6.

x² + (a + 1)x + b

∵ x = 2 is a zero

and x = – 3 is another zero

∴ (2)² + (a + 1)² + 6 = 0

and (- 3)² + (a + 1) (- 3) + b = 0

⇒ 4 + 2cr + 2 + & = 0

and 9 – 3a – 3 + b = 0

⇒ 2a + b = – 6 …(i)

and – 3a + b = – 6 …(ii)

Solving (i) and (ii), we get 5a = 0

⇒ a = 0 and b = – 6.