Question

ifalfa and bita are the zeroes of quadratic polynomial f(x)=ax^2+bx+c then calculate (1) alfa^2+bita^2 (2) alfa^2/bita +bita^2/alfa​

Answer 1

${ \bold{ \underline{ \green{ANSWER}} : - }} \\ \\ { \bold{ \underline{Given}} : - } \\ \\ { \bold{ \blue{ \: \: \: \: . \: \: a \: \: quadratic \: \: equation \: \: a {x}^{2} + bx + c = 0 }}} \\ \\ { \bold{ \underline{ To \: \: find } : - }} \\ \\ { \bold{ \blue{ \: \: \: \: \: \: (1) \: \: { \alpha }^{2} + { \beta }^{2} }}} \\ \\{ \bold{ \blue{ \: \: \: \: \: \: (2)\: \: \frac{ { \alpha }^{2} }{ \beta } + \frac{ { \beta }^{2} }{ \alpha } }}} \\ \\ { \bold{ \boxed{ \boxed{ \red{ \mathbb{ \huge \star \: \: SOLUTION \: \star }}}}}} \\ \\ { \bold{ \blue{(1) \: \: { \alpha }^{2} + { \beta }^{2} = \: ? }}} \\ \\ { \bold{ \blue{ \: \: \implies { { \alpha } }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta }}} \\ \\ { \bold{ \blue{ \: \implies \: { \alpha }^{2} + { \beta }^{2} = {( - \frac{b}{a}) }^{2} - 2 (\frac{c}{a} )} }} \\ \\ { \bold{ \blue{ \: \: \implies \: { \alpha }^{2} + { \beta }^{2} = \frac{ {b}^{2} }{ {a}^{2} } - 2 (\frac{c}{a}) }}} \\ \\ { \bold{ \blue{ (2) \: \: \: \frac{ { \alpha }^{2} }{ \beta } + \frac{ { \beta }^{2} }{ \alpha } =?}}} \\ \\ { \bold{ \blue{ \: \: \implies \frac{ { \alpha }^{2} }{ \beta } + \frac{ { \beta }^{2} }{ \alpha } = \frac{ { \alpha }^{3} + { \beta }^{3} }{ \alpha \beta } }}} \\ \\ { \bold{ \orange{ \: \: \: \: \: . \: \: we \: \: know \: \: that \: - }}} \\ \\ { \bold{ \orange{ \: \: \: {( \alpha + \beta )}^{3} = { \alpha }^{3} + { \beta }^{3} + 3 \alpha \beta ( \alpha + \beta )}}} \\ \\ { \bold{ \blue{ \: \: so \: \: that - }}} \\ \\ { \bold{ \blue{ \: \: \: \: \: \: = \frac{ {( \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta ) }{ \alpha \beta } }}} \\ \\ { \bold{ \blue{ \: \: \: \: \: \: \: = \frac{ {( - \frac{b}{a}) }^{3} - 3( \frac{c}{a})( - \frac{b}{a} ) }{ (\frac{c}{a} )} }}}$

${ \bold{ \underline{ \red{used \: \: formula}} : - }} \\ \\ { \bold{ \blue{ \: \: (1) \: \: sum \: \: of \: \: roots = \alpha + \beta = - \frac{b}{a} }}} \\ \\ { \bold{ \blue{ \: \: (2) \: \: product \: \: of \: \: roots = \alpha \beta = \frac{c}{a} }}}$