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

(i) α − β
(ii) $\frac{1}{\mathrm{\alpha }}-\frac{1}{\mathrm{\beta }}$
(iii) $\frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\beta }}-2\mathrm{\alpha \beta }$
(iv) α²β − αβ²
(v) α⁴ + β⁴
(vi) $\frac{1}{a\mathrm{\alpha }+b}+\frac{1}{a\mathrm{\beta }+b}$
(vii) $\frac{\mathrm{\beta }}{a\mathrm{\alpha }+b}+\frac{\mathrm{\alpha }}{a\mathrm{\beta }+b}$
(viii) $a\left(\frac{{\mathrm{\alpha }}^2}{\mathrm{\beta }}+\frac{{\mathrm{\beta }}^2}{\mathrm{\alpha }}\right)+b\left(\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}\right)$

(i) Given α and are the zeros of the quadratic polynomial f (x)

=

We have,

Substituting and then we get,

Hence, the value of is.

(ii) Given α and are the zeros of the quadratic polynomial f (x)

=

We have,

Substituting and then we get,

By taking least common factor we get,

Hence the value of is .

Given α and are the zeros of the quadratic polynomial f (x)

=

We have,

By cross multiplication we get,

By substituting and we get ,

Hence the value of is.

Given α and are the zeros of the quadratic polynomial f (x)

=

We have,

By taking common factor we get,

By substituting and we get ,

Hence the value of is.

Given α and are the zeros of the quadratic polynomial f (x)

=

We have,

By substituting and we get ,

By taking least common factor we get

$$\begin{gathered} {\alpha }^4+{\beta }^4={\left[{\left(\frac{-b}{a}\right)}^2-2\times \left(\frac{c}{a}\right)\right]}^2-2\times {\left(\frac{c}{a}\right)}^2 \\ ={\left[\frac{b^2}{a^2}-\frac{2c}{a}\right]}^2-2\times {\left(\frac{c}{a}\right)}^2 \\ ={\left[\frac{b^2-2ac}{a^2}\right]}^2-2\times \frac{c^2}{a^2} \\ =\frac{{\left(b^2-2ac\right)}^2}{a^4}-2\times \frac{c^2}{a^2} \\ =\frac{{\left(b^2-2ac\right)}^2-2c^2{a}^2}{a^4} \\ \mathrm{Hence}\mathrm{the}\mathrm{value}\mathrm{of}{\alpha }^4+{\beta }^4\mathrm{is}\frac{{\left(b^2-2ac\right)}^2-2c^2{a}^2}{a^4}. \end{gathered}$$

(vi) Since α and are the zeros of the quadratic polynomial

=

We have,

By substituting and we get ,

Hence, the value of is .

(vii) Since α and are the zeros of the quadratic polynomial

=

We have,

By substituting and we get ,

Hence, the value of is .

(viii) Since α and are the zeros of the quadratic polynomial

=

We have,

By substituting and we get ,

Hence, the value of is .