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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)`
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Solution
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`
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