Let α and β be the roots of the quadratic equation ax\(^{2}\) + bx + c = 0, (a ≠ 0), then the expressions of the form α + β, αβ, α\(^{2}\) + β\(^{2}\), α\(^{2}\)  β\(^{2}\), 1/α^2 + 1/β^2 etc. are known as functions of the roots α and β.
If the expression doesn’t change on interchanging α and β, then it is known as symmetric. In other words, an expression in α and β which remains same when α and β are interchanged, is called symmetric function in α and β.
Thus \(\frac{α^{2}}{β}\) + \(\frac{β^{2}}{α}\) is a symmetric function while α\(^{2}\)  β\(^{2}\) is not a symmetric function. The expressions α + β and αβ are called elementary symmetric functions.
We know that for the quadratic equation ax\(^{2}\) + bx + c = 0,
(a ≠ 0), the value of α + β = \(\frac{b}{a}\) and αβ = \(\frac{c}{a}\). To evaluate of a symmetric
function of the roots of a quadratic equation in terms of its coefficients; we
always express it in terms of α + β and αβ.
With the above information, the values of other functions of α and β can be determined:
(i) α\(^{2}\) + β\(^{2}\) = (α + β)\(^{2}\)  2αβ
(ii) (α  β)\(^{2}\) = (α + β)\(^{2}\)  4αβ
(iii) α\(^{2}\)  β\(^{2}\) = (α + β)(α  β) = (α + β) √{(α + β)^2  4αβ}
(iv) α\(^{3}\) + β\(^{3}\) = (α + β)\(^{3}\)  3αβ(α + β)
(v) α\(^{3}\)  β\(^{3}\) = (α  β)(α\(^{2}\) + αβ + β\(^{2}\))
(vi) α\(^{4}\) + β\(^{4}\) = (α\(^{2}\) + β\(^{2}\))\(^{2}\)  2α\(^{2}\)β\(^{2}\)
(vii) α\(^{4}\)  β\(^{4}\) = (α + β)(α  β)(α\(^{2}\) + β\(^{2}\)) = (α + β)(α  β)[(α + β)\(^{2}\)  2αβ]
Solved example to find the symmetric functions of roots of a quadratic equation:
If α and β are the roots of the quadratic ax\(^{2}\) + bx + c = 0, (a ≠ 0), determine the values of the following expressions in terms of a, b and c.
(i) \(\frac{1}{α}\) + \(\frac{1}{β}\)
(ii) \(\frac{1}{α^{2}}\) + \(\frac{1}{β^{2}}\)
Solution:
Since, α and β are the roots of ax\(^{2}\) + bx + c = 0,
α + β = \(\frac{b}{a}\) and αβ = \(\frac{c}{a}\)
(i) \(\frac{1}{α}\) + \(\frac{1}{β}\)
= \(\frac{α + β}{αβ}\) = b/a/c/a = b/c
(ii) \(\frac{1}{α^{2}}\) + \(\frac{1}{β^{2}}\)
= α^2 + β^2/α^2β^2
= (α + β)\(^{2}\)  2αβ/(αβ)^2
= (b/a)^2 – 2c/a/(c/a)^2 = b^2 2ac/c^2
`11 and 12 Grade Math
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