Question

If the zeroes of quadratic polynomial x 2 + (a + 1) x + b are 2 and –3, then determine the values of a and b.

Answer 1

Answer:a=0 b=-6

Step-by-step explanation:

-(a+1)/1=2+(-3)

a+1=1

a=0

b/1=2X-3

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Answer 2

Answer:

a → 0 and b → -6

Step-by-step explanation:

-[Coefficent of x / Coefficent of x^2] = Sum of zeroes

→ a - 1 / 1 = 2 + (-3)

→ a = -1 + 1

→ a = 0

and

constant term / Coefficent of x^2 = Product of Zeroes

→ b / 1 = 2*(-3)

→ b = -6

OR

x^2 + (a+1)x + b = 0 (if x = -3 or 2)

Therfore

x = 2

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

4 + 2a + 2 + b = 0

6 + 2a + b = 0

a = (-b - 6)/2 -------------- Equation 1

if x = -3

(-3)^2 + (a+1)-3 + b = 0

9 - 3a - 3 + b = 0

6 - 3a + b = 0

Putting value of A from equation 1

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

15 + 5b/2 = 0

b = (-15*2)/5

b = -6

thus

a = (-b-6)/2

a = (-{-6}-6)/2

a = 0/2 = 0

a = 0