Question

5. 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 1

GiveN :

• Quadratic polynomial is x² + (a + 1)x + b
• Zeroes of polynomial are, 2 and -3

To FinD :

• Value of a and b

SolutioN :

Let the general form of quadratic equation be, px² + qx + r. Comparing it with given equation we get,

• p = 1
• q = a + 1
• r = b

Let the zeroes be α and β , where

• α = 2
• β = -3

___________________________

Sum of zeroes (α + β) = -q/p

⇒α + β = -q/p

⇒2 + (-3) = -(a + 1)/1

⇒-1 = - (a + 1)

⇒ 1 = a + 1

⇒a = 1 - 1

⇒ a = 0

____________________

Product of zeroes (αβ) = r/p

⇒αβ = r/p

⇒2*-3 = b/1

⇒ -6 = b

⇒b = -6

_________________________

${\boxed{\sf{a\ =\ 0}}} \: \: \: \: \sf{and} \: \: \: \: \boxed{\sf{b\ =\ -6}}$

Answer 2

Answer: (d) 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