Question

if one roots of x2 + px -7 = 0 then value of p is

Answer 1

$\huge\underline \mathfrak \purple {solution}$

${ \rm{the \: value \: of \: x = 7}}$

${ \rm{ \therefore {x}^{2} + px - 7 = 0}}$

${ \rm{ \to {7}^{2} + 7p - 7 = 0}}$

${ \rm{ \to 49 + 7p - 7 = 0}}$

${ \rm{ \to 7p + 42 = 0}}$

${ \rm{ \to p = \frac{ - 42}{7} }}$

${ \rm{ \to p = - 6}}$

${ \rm{so \: the \:p \: is \: - 6}}$

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$\huge\underline \mathfrak \purple {more \: formulas}$

${ \rm{ {ax}^{2} + bx + c = 0}}$

${ \rm {b}^{2} - 4ac}$

${ \rm{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}$

${ \rm{roots \: related \: to \: it \: are}}$

${ \rm{{b}^{2} - 4ac = 0 \: root \: is \: real \: and \: equal}}$

${ \rm{{b}^{2} - 4ac > 0 \: root \: is \: real \: and \: unequal}}$

${ \rm{{b}^{2} - 4ac < 0 \: root \: is \: imaginary}}$