Question

If α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c, then evaluate :

`a(α^2/β+β^2/α)+b(α/β+β/α)`

Answer 1

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

f(x) = ax² + bx + c

∴ α + β = `(-"Coefficient of x")/("Coefficient of x"^2) = (- b/a)`

∴  αβ = `("Constant term")/("Coefficient of x"^2) = c/a`

We have,

`a(α^2/β+β^2/α)+b(α/β+β/α)`

= `a((α^3+β^3)/(αβ))+b((α^2+β^2)/(αβ))`

`= a(((α + β)^2 - 3αβ(α + β))/(αβ)) + (((α + β)^2 - 2αβ)/(αβ))  ...{(a^3 + b^3 = (a + b)^3 - 3ab(a + b)),(a^2 + b^2 = (a + b)^2 - 2ab):}`

By substituting α + β = `(-b)/a` and αβ = `c/a`, we get ,

= `a[((- b/a)^3 - 3c/a(-b/a))/(c/a)] + b[((-b/a)^2 - 2c/a)/(c/a)]`

= `a[(-b^3/a^3 + (3bc)/a^2)/(c/a)] + b[(b^2/a^2 - (2c)/a)/(c/a)]`

= `a[((-b^3 + 3abc)/a^3)/(c/a)] + b[((b^2 - 2ac)/(a^2))/(c/a)]`

= `a[(-b^3 + 3abc)/a^3 × a/c] + b[(b^2 - 2ac)/a^2 × a/c]`

= `a[(-b^3 + 3abc)/(a × a × cancel(a)) × cancel(a)/c] + b[(b^2 - 2ac)/(a × cancel(a)) × cancel(a)/c]`

= `a[(-b^3 + 3abc)/(a × a × c)] + b[(b^2 - 2ac)/(ac)]`

= `(cancela(-b^3 + 3abc))/(cancela × a × c) + (b(b^2 - 2ac))/(a × c)`

= `(-b^3 + 3abc)/(ac) + (b^3 - 2abc)/(ac)`

= `(cancel(-b^3) + 3abc + cancel(b^3) - 2abc)/(ac)`

= `(cancelabcancelc)/(cancelacancelc)`

= b

`a(α^2/β+β^2/α)+b(α/β+β/α) = b`