Question

If α and β are the zeros of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate: (vi)1/(αα+b)+1/(αβ+b)​

Answer 1

Given info : If α and β are the zeroes of the quadratic polynomial f(x) = ax² + bx + c.

To find : The value of 1/(aα + b) + 1/(aβ + b) is ...

solution : α and β are zeroes of the polynomial f(x) = ax² + bx + c

so, f(α) = aα² + bα + c = 0

⇒α(aα + b) = -c

⇒α/-c = 1/(aα + b) ...(1)

f(β) = aβ² + bβ + c = 0

⇒β(aβ + b) = -c

⇒β/-c = 1/(aβ + c) ...(2)

adding equations (1) and (2) we get,

⇒α/-c + β/-c = 1/(aα + b) + 1/(aβ + b)

⇒1/(aα + b) + 1/(aβ + b) = -(α + β)/c

sum of products = α + β = -b/a

⇒1/(aα + b) + 1/(aβ + b) = -(-b/a)/c = b/ac

Therefore the value of 1/(aα + b) + 1/(aβ + b) is b/ac