Question

Let f(x) be a quadratic polynomial such that f(-1) + f(2) =0. If one of the roots of f(x) = 0 is 3,then its other roots lies in: choices are (1) (-3,-1) (2) (1,3) (3)( -1,0) (4) (0,1)​

Answer 1

Let the quadratic polynomial be ax² + bx + c and its another root be k.

Hence, f(x) = ax² + bx + c

Concept

• Product of roots = $\dfrac{c}{a}$

Calculations

Given, f(3) = 0 .

Hence, a(3)² + b(3) + c = 0

=> 9a + 3b + c = 0..........eqn(i)

Also,given f(-1) + f(2) = 0

=> a(-1)² + b(-1) + c + a(2)² + b(2) + c = 0

=> a - b + c + 4a + 2b + c = 0

=> 5a + b + 2c = 0

Multiplying by 3 , we get

=> 15a + 3b + 6c = 0 .....eqn(ii)

Subtracting eqn(i) from (ii)

=> (15a + 3b + 6c ) - (9a + 3b + c) = 0

=> 6a + 5c = 0

=> 5c = -6a

=> $\dfrac{c}{a} \:=\: \dfrac{-6}{5}$

Now, product of roots = $\dfrac{c}{a}$

=> 3k = $\dfrac{-6}{5}$

=> k = $\dfrac{-6}{5 \times 3}$

=> k = $\dfrac{-6}{15}$

The other root is $\dfrac{-6}{15}$ which lies between ( -1,0).

Hence, the correct option is (3).