Question

If α and β are the zeroes of the quadratic polynomial f(x) = ax² + bx + c, then evaluate `1/(aalpha+b)+1/(abeta+b)`.

Answer 1

Since α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c

α + β = `-"Coefficient of x"/"Coefficient of x"^2`

= `(-b)/a`

ab = `"Constant term"/"Coefficient of x"^2`

= `c/a`

We have, `1/(aalpha + b) + 1/(abeta + b)`

`1/(aalpha + b) + 1/(abeta + b) = (abeta + b + aalpha + b)/((aalpha + b)(abeta + b))`

`1/(aalpha + b) + 1/(abeta + b) = (a(alpha + beta) + 2b)/(a^2 xx alphabeta + ab beta + ab alpha + b^2)`

`1/(aalpha + b) + 1/(abeta + b) = (a(alpha + beta) + 2b)/(a^2 alpha beta + ab(alpha + beta)+ b^2)`

By substituting `a + beta = (-b)/a and alphabeta = c/a "we get",`

`1/(aalpha + b) + 1/(abeta + b) = (a xx(-b)/a + 2b)/(a^2 xx c/a + ab xx (-b)/a + b^2)`

`1/(aalpha + b) + 1/(abeta + b) = (cancela xx (-b)/cancela + 2b)/(a^(cancel2^1)xx c/cancela + cancelab xx (-b)/cancela + b^2)`

`1/(aalpha + b) + 1/(abeta + b) = (-b + 2b)/(a xx c - b^2 + b^2)`

`1/(aalpha + b) + 1/(abeta + b) = b/(ac - cancel(b^2) + cancel(b^2))`

`1/(aalpha + b) + 1/(abeta + b) = b/(ac)`

Hence, the value of `1/(aalpha+b)+1/(abeta+b)  "is"  b/(ac)`.