Question

If alpha and beta are the zeros of the polynomial ax2+bx+c then evaluate alpha/beta+beta/alpha

Answer 1

Polynomial

        ↓

• α + β = - Coefficient of x/Coefficient of x²

• αβ = Constants term/Coefficient of x²

• (a + b)² = a² + b² + 2ab

$p(x) = a{x}^{2} + bx + c \\ \\ \alpha + \beta = \frac{ - b }{a} \\ \alpha \beta = \frac{c}{a} \\ {( \alpha + \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta \\ \frac{ {b}^{2} }{ {a}^{2} } = { \alpha }^{2} + { \beta }^{2} + 2 \frac{c}{a} \\ \frac{ {b}^{2} - 2ac }{ {a}^{2} } = { \alpha }^{2} + { \beta }^{2} \\ \\ Now \:\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } \\ \\ \frac{ {b}^{2} - 2ac}{ {a}^{2} } \times \frac{a}{c} = \bold{\frac{ {b}^{2} - 2ac }{ac}}$

Answer 2

$\huge{\underline{\bf Answer:}}$

Polynomial = ax² + bx + c

Sum of zeroes (α + β) = -b/a

Product of zeroes (αβ) = c/a

$( \alpha + \beta ) {}^{2} = \alpha {}^{2} + \beta {}^{2} + 2 \alpha \beta \\ \\ (\frac{ - b}{a} ) {}^{2} = \alpha {}^{2} + \beta {}^{2} + 2 \times \frac{c}{a} \\ \\ \frac{b {}^{2} }{a{}^{2} } - \frac{2c}{a} = \alpha {}^{2} + \beta {}^{2} \\ \\\alpha {}^{2} + \beta {}^{2} = \frac{b {}^{2} - 2ac }{a {}^{2}}\\ \\ \frac{ \alpha }{ \beta}+\frac{\beta}{ \alpha}\\ \\ \frac{ \alpha {}^{2} + \beta ^{2}}{ \alpha \beta}\\ \\\frac{b {}^{2} - 2ac }{a {}^{2}} \times \frac{a}{c} \\ \\ \boxed{\bf \frac{b^{2}-2ac}{ac}}$