Question

Please solve the 35Q in the attached image

Answer 1

Given that 3 is a root of f(x) = ax² + bx + c = 0.

f(3) = a(3)² + b(3) + c = 0

f(3) = 9a + 3b + c = 0

4 · f(3) = 4(9a + 3b + c) = 4 · 0

4 · f(3) = 36a + 12b + 4c = 0     →     (1)

Given that  f(5) = - 3 · f(2).

    f(5) = - 3 · f(2)

⇒  a(5)² + b(5) + c = - 3 (a(2)² + b(2) + c)

⇒  25a + 5b + c = - 3(4a + 2b + c)

⇒  25a + 5b + c = - 12a - 6b - 3c

⇒  25a + 5b + c + 12a + 6b + 3c = 0

⇒  37a + 11b + 4c = 0     →     (2)

From (1) and (2),

    37a + 11b + 4c = 36a + 12b + 4c

⇒  37a + 11b + 4c - 36a - 12b - 4c = 0

⇒  a - b = 0

⇒  a = b

⇒  b / a = 1

⇒ - b / a = - 1

From (1),

    36a + 12b + 4c = 0

⇒  36a + 12a + 4c = 0 = 36b + 12b + 4c

⇒  48a + 4c = 0 = 48b + 4c

⇒  4(12a + c) = 0 = 4(12b + c)

⇒  12a + c = 0 = 12b + c

⇒  a = b = - c / 12

⇒  c = - 12a = - 12b

Now,

(A)  Let the other root be k.

Sum of roots = k + 3 = - b / a = - 1

k + 3 = - 1  ⇒  k = - 1 - 3 = - 4.

Hence option (2) is the answer.

(B)  In terms of a,

a + b + c  =  a + a - 12a = - 10a

In terms of b,

a + b + c = b + b - 12b = - 10b

In terms of c,

a + b + c = - c / 12  - c / 12  + c  =  5c / 6

From each of these, we can find out that the value of  a + b + c  is dependent on each coefficient, means it won't be a constant. So we can't determine the value of it.

Hence option (5) is the answer.