Question

please solve correct answer will explanation I will mark you brainliest.​​

Answer 1

$\huge\red{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{Option \ (a)-1 \ is \ a \ right \ answer.}$

$\sf{The \ value \ of \ a \ is \ -1.}$

$\sf\orange{Given:}$

$\sf{The \ quadratic \ equations \ are}$

$\sf{\implies{x^{2}-ax-6=0}}$

$\sf{\implies{x^{2}+ax-2=0}}$

$\sf\pink{To \ find:}$

$\sf{The \ value \ of \ a.}$

$\sf\green{\underline{\underline{Solution:}}}$

(a)$\sf{Let \ substitute \ a=-1 \ in \ both \ equations}$

(1)$\sf{\implies{x^{2}-(-1)x-6=0}}$

$\sf{\implies{x^{2}+1x-6=0}}$

$\sf{\implies{x^{2}+3x-2x-6=0}}$

$\sf{\implies{x(x+3)-2(x+3)=0}}$

$\sf{\implies{(x+3)(x-2)=0}}$

$\sf{\implies{x=-3 \ or \ 2}}$

(2)$\sf{\implies{x^{2}+(-1)x-2=0}}$

$\sf{\implies{x^{2}-x-2=0}}$

$\sf{\implies{x^{2}+x-2x-2=0}}$

$\sf{\implies{x(x+1)-2(x+1)=0}}$

$\sf{\implies{(x+1)(x-2)=0}}$

$\sf{\implies{x=-1 \ or \ 2}}$

$\sf{Here \ 2 \ is \ a \ common \ root \ in}$

$\sf{in \ both \ equations.}$

$\sf\purple{\tt{\therefore{The \ value \ of \ a \ is \ -1}}}$