Question

5. If c and B are zeros of p(x) X2 - x - 30 then ? + B
O (A) 91
O (B) 125
O (C) 216
O (D) 341​

Answer 1

AS I UNDERSTAND UR QUESTION :-

GIVEN :

$\sf \: c \: and \: b \: \: are \: \: the \: zeros \: \: of \: \: the \: \: polynomial$

$\sf \: P(x) = {x}^{2} - x - 30$

TO FIND :-

$\sf \: value \: of \: c + b =$

SOLUTION :-

$\sf If \: zeros \: are \: given \: it \: means \: the \: value \: of \: polynomial \\ \sf \: at \: these \: become \: zero \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: At \: \: x =c \: \:putting \: first \: root \: in \: given \: polynomial.$

$\sf \: P(c) = {c}^{2} - c - 30$

$\sf \: Here \:( c) \: is \:zero \: of \: the \: polynomial \: then$

$\sf \: given \: polynomial \: become \: zero$

$\sf \implies {c}^{2} - c - 30 = 0$

$\sf \implies {c}^{2} -6 c + 5c - 30 = 0$

$\sf \implies c(c -6) + 5(c - 6) = 0$

$\sf \implies (c -6) (c + 5) = 0$

$\sf \: Then \: c = 6 \: or \: \: - 5$

$\large \sf \: Similarly : -$

$\sf \: At \: \: x =b \: \:putting in \: given \: polynomial.$

$\sf \: P(c) = {b}^{2} - b - 30$

$\sf \implies {b}^{2} -6 b + 5b - 30 = 0$

$\sf \implies b(b -6) + 5(b - 6) = 0$

$\sf \implies (b -6) (b + 5) = 0$

$\sf \: Then \: b = \: 6 \: \: \: \: or \: \: - 5$

$\sf \: If \: \: given \: \: question \: is \: \: (c+b )$

$\sf \: Then \: \: it \: \: may \: \: be$

$\sf \: (c + b) = 6 + 6 = 12$

$\sf \: (c + b) = 6 + ( - 5) = 1$

$\sf \:( c + b) = - 5 + ( - 5 )= - 10$