Question

hey !! help me!!

full explanation !

Answer 1

heya..!!!!!


given that:-

the zeroes of the Quadratic polynomial
px² + qx + r, (r not equal to 0)

D = b² - 4ac = 0

=> q² - 4pr = 0

=> q² = 4pr

since,L.H.S (q²) can never be negative

AND,

also R.H.S (4pr) can never be negative

then,

(p and r must be same sign)

if p and r have opposite sign, then p and r will be negative sign which is never possible

HENCE, option (3) is correct

Answer 2

$The \: \: answer \: \: is \: \: given \: \: below.\\ \\ Now, \: \: the \: \: polynomial \: \: is \\ given \: \: by \\( p {x}^{2} + qx + r). \\ \\ Using \: \: Sridhar \: \: Acharya's \: \: \\formula, \:\: we \: \: get \\ \\ x = \frac{ - q + \sqrt{ {q}^{2} - 4pr} }{2p} \\ \\ and \\ \\ x = \frac{ - q - \sqrt{ {q}^{2} - 4pr} }{2p} \\ \\ For \: \: equal \: \: zeroes, \: \: we \\ must \: \: have \\ \\ the \: \: discriminant = 0 \\ \\ So, \: \: {q}^{2} - 4qr =0\\ \\ Or, \: \: {q}^{2} = 4pr \\ \\ Since, \: \: {q}^{2} \: \: is \: \: always \: \: positive \\ 4pr \: \: can \: \: not \: \: be \: \: negative. \\ \\ Thus \: \: the \: \: only \: \: possibility \\ for \: \: that \: \: is \: \: p \: \: and \: \: r \: \: must \\ have \: \: same \: \: sign, \\ either \: \: both \: \: negative \: \: or \\ both \: \: positive. \\ \\ Hence, \: \: option \: \: (3) \: \: is \: \: correct. \\ \\ Thank \: \: you \: \: fo r\: \: your \: \: question.$