Question

I
If a, ß are the roots of ax2+bx+C=0,then
$\alpha \beta 2 + \alpha 2 \beta + \alpha \beta \: is$
​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{{ \alpha }^{2} + { \beta }^{2} + \alpha \beta = \frac{ {b}^{2} - ac }{ {a}^{2} }}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies {ax}^{2} + bx + c = 0 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: roots \: of \: the \:eqn \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \tt: \implies { \alpha }^{2} + { \beta }^{2} + \alpha \beta = ?$

• According to given question :

$\green {\star} \tt \: a = a \: \: \: \: \: \: b = b \: \: \: \: \: \: c = c \\ \\ \bold{As \:we \: know \: that} \\ \tt: \implies Sum \: of \: roots = - \frac{ b}{a} \\ \\ \tt: \implies \alpha + \beta = - \frac{b}{a} - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Product \: of \: roots = \frac{c}{a} \\ \\ \tt: \implies \alpha \beta = \frac{c}{a} - - - - - (2) \\ \\ \bold{For \: finding \: values } \\ \tt: \implies { \alpha }^{2} + { \beta }^{2} + \alpha \beta \\ \\ \tt: \implies { (\alpha + \beta )}^{2} - 2 \alpha \beta + \alpha \beta \\ \\ \tt: \implies { (\alpha + \beta )}^{2} - \alpha \beta \\ \\ \tt: \implies \bigg(- \frac{b}{a} \bigg)^{2} - \frac{c}{a} \\ \\ \tt: \implies \frac{ {b}^{2} }{ {a}^{2} } - \frac{c}{a} \\ \\ \tt: \implies \frac{ {b}^{2} - ac}{ {a}^{2} } \\ \\ \green{\tt \therefore { \alpha }^{2} + { \beta }^{2} + \alpha \beta = \frac{ {b}^{2} - ac }{ {a}^{2} } }$

Answer 2

$\huge\sf\underline\red{Refer the attachment}$