Question

if -2 and 3 are the zeroes of the quadratic polynomial x^2 + (a+b)x+b, then the values of a and b are?

Answer 1

Step-by-step explanation:

Given -

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

To Find -

• Value of a and b

Now,

→ p(x) = x² + (a + b)x + b

→ p(-2) = (-2)² + (a + b)-2 + b

→ 4 - 2a - 2b + b = 0

→ 4 = 2a + b ...... (i)

And

p(3) = (3)² + (a + b)3 + b

→ 9 + 3a + 3b + b = 0

→ 3a + 4b = -9 ..... (ii)

From (i) and (ii), we get :-

→ [ 2a + b = 4 ] × 4

[ 3a + 4b = -9 ] × 1

→ 8a + 4b = 16

3a + 4b = -9

(-) (-) (+)

____________

→ 5a = 25

→ a = 25/5

→ a = 5

Now,

Substituting the value of a on 2a + b = 4

→ 2(5) + b = 4

→ b = 4 - 10

→ b = -6

Hence,

The value of a is 5 and b is -6.