Question

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

Answer 1

f(x) = ax² + bx + c

α + β = `(-b/a)`

αβ = `c/a`

since α + β are the roots (or) zeroes of the given polynomials

then

`beta/(aalpha+b)+alpha/(abeta+b)`

`=(beta(abeta+b)+alpha(aalpha+b))/((aalpha+b)(abeta+b))`

`=(abeta^2+b beta+aalpha^2+balpha)/(a^2alphabeta+abalpha+ab beta+b^2)`

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

`=(a(alpha^2+beta^2)+b(alpha+beta))/(ac+ab(-b/a)+b^2)`

`=(a[(alpha+beta)^2-2alphabeta]+bxx-b/a)/(ac-b^2+b^2)`

`=(a[(-b/a)^2-2(c/a)]-b^2/a)/(ac)`

`=((b^2)/a-(2c)-b^2/a)/(ac)`

`=(-2c)/(ac)`

`=(-2)/a`