Question

4) If the zeroes of the quadratic polynomial x2 + (a + 1)x + b are 2 and –3, then find the value of a​

Answer 1

Method to be used:

• Put the value of x, first as 2 and then as 3. Then, equate the equations to 0 [Since they are the zeroes, they'll give the result as 0]
• Obtain two equations with variables a and b. Solve it with any which method you like.

Solution:

When x = 2

p(x) = x^2 + (a + 1)x + b

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

p(2) = 4 + 2a + 2 + b = 0

p(2) => b = -6 - 2a __(1)

When x = -3

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

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

p(-3) = 9 - 3a - 3 + b = 0

Put (1):

6 - 3a - 6 - 2a = 0

=> a = 0

b = -6 - 2a

=> b = -6

Answer 2

Solutions:

x = 2

f(x) = x² + (a + 1)x + b

(2)² + (a + 1)2 + b =0

4 + 2a + 2 + b = 0

b = -b - 2a

x = -3

x² + (a + 1)x + b

(-3)² + (a + 1)-3 + b =0

9 - 3a - 3 + b = 0

6 - 3a - 6 - 2a = 0

a = 0

b = -6 - 2a

b = -6.

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