Question

Q1.If α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c, then evaluate:
(i) α – β (ii) (1/α) – (1/β)
(iii) (1/α) + (1/β) – 2αβ (iv) α²β + αβ²
(v) α⁴ + β⁴ (vi) 1/(aα + b) + 1/(aβ + b)
(vii) β/(aα + b) + α/(aβ + b) (viii) a[(α²/β) + (β²/α)] + b[(α/β) + (β/α)]

Answer 1

Step-by-step explanation:

Given polynomial, f(x) = ax² + bx + c

If, α and β are the zeros of the quadratic polynomial.

$\begin{gathered} \alpha + \beta = \frac{ - b}{a} \\ \\ \alpha \beta = \frac{c}{a} \end{gathered}$

Now, (1)

$\begin{gathered}( \alpha - \beta )^{2} = ( \alpha + \beta ) ^{2} - 4 \alpha \beta \\ \\ ( \alpha - \beta ) ^{2} = ( \frac{ - b}{a} ) ^{2} - 4 (\frac{c}{a} ) \\ \\ \alpha - \beta = \frac{\sqrt{ {b}^{2} - 4ac} }{a}\end{gathered}$

Now (2)

$\begin{gathered} \frac{1}{ \alpha } - \frac{1}{ \beta } = \frac{ \beta - \alpha }{ \alpha \beta } = \frac{ - \sqrt{ {b}^{2} - 4ac} \times a }{c \times a} \\ \\ \frac{1}{ \alpha } - \frac{1}{ \beta } = \frac{ - \sqrt{ {b}^{2} - 4ac}}{c} \end{gathered}$

Now (3)

$\begin{gathered} \frac{1}{ \alpha } + \frac{1}{ \beta } - 2 \alpha \beta \\ \\ = \frac{ \alpha + \beta }{ \alpha \beta } - 2 \alpha \beta \\ \\ = \frac{ - b}{c} - 2( \frac{c}{a} ) \\ \\ = - \frac{ab - {c}^{2} }{ac} \\ \\ = \frac{ {c}^{2} - ab}{ac} \end{gathered}$

Now (4)

$\begin{gathered} \alpha^{2} \beta + \alpha \beta ^{2} \\ \\ = \alpha \beta ( \alpha + \beta ) \\ \\ = ( \frac{ - b}{a} )( \frac{c}{a} ) \\ \\ = \frac{ - bc}{a ^{2} } \end{gathered}$

Now (5)

$\begin{gathered} { \alpha }^{4} + { \beta }^{4} \\ \\ = ( \alpha ^{2} + \beta ^{2} ) ^{2} - 2 { \alpha }^{2} { \beta }^{2} \\ \\ =( ( \alpha + \beta )^{2} - 2 \alpha \beta ) ^{2} - 2( { \alpha \beta })^{2} \\ \\ = {( (\frac{b ^{2} }{ {a}^{2} }) - 2 (\frac{c}{a}) ) }^{2} - 2( \frac{ {c}^{2} }{ {a}^{2} }) \\ \\ = (\frac{ {b}^{2} - 2ac }{ {a}^{2} } ) ^{2} - 2 \frac{ {c}^{2} }{ { a}^{2} } \\ \\ = \frac{ {b}^{4} + 4 {a}^{2} {c}^{2} - 4ac {b}^{2} - 2 {a}^{2} {c}^{2} }{ {a}^{4} } \\ \\ = \frac{ {b}^{2}(b ^{2} - 4ac) + 2 {a}^{2} {c}^{2} }{ {a}^{4} } \end{gathered}$

Now (6)

$\begin{gathered} \frac{1}{a \alpha + b} + \frac{1}{a \beta + b} \\ \\ = \frac{a( \alpha + \beta ) + 2b}{( a\alpha + b)( a \beta + b)} \\ \\ = \frac{ - b + 2b}{ {a}^{2}( \frac{c}{a} ) + {b}^{2} + ab( \frac{ -b }{a} )} \\ \\ = \frac{b}{ac + {b}^{2} - {b}^{2} } \\ \\ = \frac{b}{ac} \end{gathered}$

Now (7)

$</p><p>\begin{gathered} \frac{ \beta }{a \alpha + b} + \frac{ \alpha }{a \beta + b} \\ \\ = \frac{ \beta (a \beta + b) + \alpha (a \alpha + b)}{(a \alpha + b)(a \beta + b)} \\ \\ = \frac{a( { \alpha }^{2} + { \beta }^{2} ) + b( \alpha + \beta )}{ {a}^{2} } \\ \\ = \frac{a( \frac{b ^{2} }{ {a}^{2} } - \frac{2c}{a}) - \frac{ {b}^{2} }{a} }{ {a}^{2} } \\ \\ = \frac{ \frac{ {b}^{2} - 2ac - {b}^{2} }{a} }{ {a}^{2} } \\ \\ = \frac{ - 2c}{ {a}^{2} } \\ \end{gathered}$

Now (8)

$\begin{gathered}a( \frac{ { \alpha }^{2} }{ \beta } + \frac{ { \beta }^{2} }{ \alpha } ) + b( \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } ) \\ \\ = \frac{a( { \alpha }^{3} + { \beta }^{3}) + b( { \alpha }^{2} + { \beta }^{2} ) }{ \alpha \beta } \\ \\ = \frac{a( \frac{ {b}^{3} }{ {a}^{3} } + 3 \frac{bc}{ {a}^{2} } ) + b( \frac{ {b}^{2} }{ {a}^{2} } - 2 \frac{c}{a}) }{ \alpha \beta } \\ \\ = \frac{ {b}^{3} + 3abc + {b}^{3} - 2abc }{ {a}^{2} \times \frac{c}{a} } \\ \\ = \frac{2 {b}^{3} + abc }{ac} \end{gathered}$