Question

(v). If the sum of the roots is –p and product of the roots is –1/p, then the quadratic polynomial is
a) k(–px
2
+ x/p + 1)
b) k(px2
– x/p – 1)
c) k(x2
+ px – 1/p)
d) k(x2
– px + 1/p)

Answer 1

Answer:

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Step-by-step explanation:

$so \: for \: a \: certain \: quadratic \: polynomial \: \\ let \: its \: two \: roots \: be \: \alpha \: and \: \beta \:$

$so \: here \: \: \alpha + \beta = - p \\ \alpha \beta = –1/p$

$so \: we \: know \: that \\ general \: formula \: of \: any \: quadratic \: polynomial = x.square - ( \alpha + \beta )x + \alpha \beta = 0$

$ie \: substitute \: given \: values \: in \: it \\ we \: get \: then \\ x.square - ( - p)x + (–1/p) = 0 \\ x.square + px \: –1/p = 0 \\ \\ multiplying \: througout \: by \: p \\ we \: get \\ px.square +p.square \: x - 1 = 0$

$hence \: the \: required \: quadratic \: polynomial \: is \\ px.square + p.square \: x - 1 = 0$