Question

find the zeros of the polynomial and verify the relationship between the zeros and their cofficients of
xsquare+x-12​

Answer 1

Step-by-step explanation:

hope this helps you............. ..

Answer 2

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{Zeroes \ are \ -4 \ and \ 3.}$

$\bold\orange{Given:}$

$\bold{The \ given \ polynomial \ is}$

$\bold{=>x^{2}+x-12}$

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

$\bold{Zeroes \ of \ the \ polynomial.}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{The \ given \ polynomial \ is}$

$\bold{=>x^{2}+x-12}$

$\bold{Here, \ a=1, \ b=1 \ and \ c=-12}$

$\bold{=>x^{2}+4x-3x-12}$

$\bold{=>x(x+4)-3(x+4)}$

$\bold{=>(x+4)(x-3)}$

$\bold{\tt{\therefore{x=-4 \ or \ 3}}}$

$\bold\purple{\tt{\therefore{Zeroes \ are \ -4 \ and \ 3.}}}$

__________________________________

$\bold\blue{Verification}$

$\bold{Let \ \alpha \ be \ -4 \ and \ \beta \ be \ -4}$

$\bold{Sum \ of \ zeroes=\frac{-b}{a}}$

$\bold{Product \ of \ zeroes=\frac{c}{a}}$

___________________________________

$\bold{\alpha+\beta=-4+3}$

$\bold{\therefore{\alpha+\beta=-1...(1)}}$

$\bold{\frac{-b}{a}=\frac{-(1)}{1}}$

$\bold{\therefore{\frac{-b}{a}=-1...(2)}}$

$\bold{from \ (1) \ and \ (2)}$

$\bold{Sum \ of \ zeroes=\frac{-b}{a}}$

___________________________________

$\bold{\alpha×\beta=-4×3}$

$\bold{\therefore{\alpha×\beta=-12...(3)}}$

$\bold{\frac{c}{a}=\frac{12}{1}}$

$\bold{\therefore{\frac{c}{a}=12...(4)}}$

$\bold{from \ (3) \ and (4)}$

$\bold{Product \ of \ zeroes=\frac{c}{a}}$

$\bold{Hence, \ verified.}$