If α and β are the zeros of the quadratic polynomial

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

(i)  \alpha^2 + \beta^2

(ii)  \dfrac{1}{\alpha}+\dfrac{1}{\beta}

(iii) \dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}

(iv) \alpha^3 + \beta^3

(v)  \dfrac{1}{\alpha^3}+\dfrac{1}{\beta^3}

(vi) \dfrac{\alpha^2}{\beta}+\dfrac{\beta^2}{\alpha}

Solution:  Since α and β are the zeros of the quadratic polynomial

f(x) = ax² + bx + c

\alpha + \beta = -\dfrac{b}{a} and \alpha . \beta = \dfrac{c}{a}

(i) We know that

(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha . \beta

\Rightarrow \alpha^2 + \beta^2 = (\alpha +\beta)^2 - 2\alpha . \beta

\Rightarrow \alpha^2 + \beta^2 = (-\dfrac{b}{a})^2 - 2(\dfrac{c}{a})

\Rightarrow \alpha^2 + \beta^2 = \dfrac{b^2}{a^2} - 2(\dfrac{c}{a}) = \dfrac{b^2 - 2ac}{a^2}

(ii) we have,

\dfrac{1}{\alpha}+\dfrac{1}{\beta}

=\dfrac{\beta + \alpha}{\alpha . \beta}

=\dfrac{-\frac{b}{a}}{\frac{c}{a}}

= -\dfrac{b}{c}

(iii) We have,

\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}

=\dfrac{\alpha^2 + \beta^2}{\alpha . \beta}

=\dfrac{\frac{b^2 - 2ac}{a^2}}{\frac{c}{a}} [ From (ii) ]

=\dfrac{b^2 - 2ac}{ac}

(iv) We know that

(\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha .\beta(\alpha + \beta)

\Rightarrow \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha. \beta (\alpha +\beta)

\Rightarrow \alpha^3 + \beta^3 = (-\frac{b}{a})^3 - 3(\frac{c}{a})(-\frac{b}{a})

\Rightarrow \alpha^3 + \beta^3 = -\frac{b^3}{a^3} + 3\frac{bc}{a^2}

\Rightarrow \alpha^3 + \beta^3 = \dfrac{-b^3 + 3abc}{a^3} = \dfrac{3abc-b^3}{a^3}

(v) We have,

\dfrac{1}{\alpha^3}+\dfrac{1}{\beta^3}

=\dfrac{\alpha^3+\beta^3}{\alpha^3. \beta^3}

=\dfrac{\alpha^3+\beta^3}{(\alpha. \beta)^3}

=\dfrac{\frac{3abc-b^3}{a^3}}{(\frac{c}{a})^3} [From (iv) ]

=\dfrac{3abc - b^3}{c^3}

(vi) \dfrac{\alpha^2}{\beta}+\dfrac{\beta^2}{\alpha}

= \dfrac{\alpha^3 + \beta^3}{\alpha. \beta}

= \dfrac{(\alpha + \beta)^3 - 3\alpha . \beta(\alpha + \beta)}{\alpha . \beta}

= \dfrac{-\frac{b^3}{a^3} + 3\frac{bc}{a^2}}{\frac{c}{a}}

= \dfrac{3abc-b^3}{a^2c}

Some other question:

1: If we add 1 to the numerator and subtract 1 from the

2:Find the zeros of the quadratic polynomial x² + 7x + 12

3: find the value of k for which the pair of equation

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