Question

if alpha and beta are the zeros of quadratic polynomial p of x²-5x+6 find beta upon alpha plus alpha upon beta and alpha -beta​

Answer 1

$\bf \: Given$

$\sf \to \: {x}^{2} - 5x + 6 = 0$

$\bf \: To \: find$

$\sf \to \: \dfrac{ \beta }{ \alpha } + \dfrac{ \alpha }{ \beta } \: and \: \alpha - \beta$

$\bf \:Now \: Compare \: with$

$\sf \to \: a {x}^{2} + bx + c = 0$

$\bf \: We \: get$

$\sf \to \: a = 1,b = - 5 \: and \: c \: = 6$

$\bf \: We \: know \: that$

$\sf \to \: \alpha + \beta = \dfrac{ - b}{a} = \dfrac{ - ( - 5)}{1} = 5$

$\sf \to \alpha \beta = \dfrac{c}{a} = \dfrac{6}{1} = 6$

$\bf \: Now \: Take$

$\sf \to \: \dfrac{ \beta }{ \alpha } + \dfrac{ \alpha }{ \beta }$

$\bf Now \: Simplify \: the \: equation$

$\sf \to \: \dfrac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta }$

$\sf \to \: \dfrac{( \alpha + \beta )^2 - 2 \alpha \beta }{ \alpha \beta }$

$\bf \: Now \: Put \: the \: value$

$\sf \to \: \alpha + \beta = 5 \\ \sf \to \alpha \beta = 6 \: \: \: \: \: \:$

$\bf \: We \: get$

$\sf \to \: \dfrac{25 - 2 \times 6}{6} = \dfrac{25 - 12}{6} = \dfrac{ 13}{6}$

$\bf \: Now \: Take$

$\sf \to \alpha - \beta$

$\bf \: Simplify \: the \: equation$

$\sf \to \sqrt{( \alpha - \beta ) {}^{2} }$

$\sf \to \sqrt{ { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta }$

$\sf \to \sqrt{( \alpha + \beta ) {}^{2} - 2 \alpha \beta - 2 \alpha \beta }$

$\sf \to \: \sqrt{( \alpha + \beta ) {}^{2} - 4 \alpha \beta }$

$\bf \: Now \: Put \: the \: value$

$\sf \to \: \alpha + \beta = 5 \\ \sf \to \alpha \beta = 6 \: \: \: \: \: \:$

$\sf \to \: \sqrt{(5) { }^{2} - 4 \times 6}$

$\sf \to \sqrt{25 - 24} = \sqrt{1} = 1$

$\bf \:Answer$

$\sf \to \: \dfrac{ \beta }{ \alpha } + \dfrac{ \alpha }{ \beta } = \dfrac{ 13}{6} \: and \: \alpha - \beta = 1$

Answer 2

Hlo this might be the answer tried my best