Question

plz tell this thing is a=/</>b​

Answer 1

Answer:

please ask this question on google you will get correct answer

Answer 2

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$\huge{\mathfrak{\purple{\boxed{\orange{\underbrace{\overbrace{\green{‡Answer}}}}}}}}$

$\: \: \: \: \: \star \displaystyle \underline {\boxed{ \bf \red{as \: we \: know \: that}}}$

$\: \: \: \blue \star \displaystyle \underline{ \boxed{ \bf{quadratic \: equation}}} \to \displaystyle \sf {(1)the \: equation \: having \: two \: roots \: (2)it \: is \: in \: the \: form \: of \: {ax}^{2} + bx + c = 0}$

$\: \: \ \: \ \ \ \sf{(3) for \: finding \: the \: value \: of \: discriminant \: we \: use \: formula \: ( {b}^{2} - 4ac)}$

$\: \: \: \bigstar \displaystyle \boxed{ \bf\red{ \: properties \:of \: discriminant \: form}}$

$\: \: \: \bullet \rm{ {b}^{2} - 4ac > 0 \: \: roots \: are \: real \: and \: distinct}$

$\: \: \: \bullet \rm{ {b}^{2} - 4ac > 0 \: \: \: \: roots \: are \: real \: and \: unequal}$

$\: \: \: \bullet \rm{ {b}^{2} - 4ac = 0 \: roots \: are \: imaginary}$

$\: \: \pink \bigstar \underline {\boxed { \bf \red{ \: formula \: method}}}$

$\: \: \: \implies \boxed{ \sf{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}$

$\: \: \: \: \dashrightarrow : \sf{ \alpha \: and \beta \: are \: the \: roots \: of \: the \: quadratic \: equation}$

$\: \: \: \: \bullet \bf{ \alpha = \frac{ - b + \sqrt {{b}^{2} - 4ac}}{2a} } \: \: \: ........... \alpha > \beta$

$\: \: \: \: \bullet \bf { \beta = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} } \: \: \: \: ....... \alpha < \beta$

$\: \: \: \: \mapsto \boxed { \tt \pink{if \: \alpha= \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \: \: then \: \alpha > \beta }}$

$\: \: \: \bigstar \mathfrak{hence \: proved}$