Question

find the zeroes snd cofficients of
$2 {x}^{2} -8x + 6$
and verify the relationship between the zeroes and the cofficients ​

Answer 1

$\Huge \bf \green{Solution:-}$

$\bf 2 {x}^{2} -8x + 6$

By Splitting the middle Term

$\bf 2x^{2} -6x-2x + 6=0$

$\bf 2x(x-3) -2(x - 3)=0$

$\bf (x-3) (2x - 2)=0$

$\bf (x-3) =0 \:\:Or\:\:2x-2=0$

$\bf x=3 \:\:Or\:\:x=\dfrac{2}{2}$

$\bf x=3 \:\:Or\:\:x=1$

$\Large\bf\red{Verification:-}$

LET,

$\bf \alpha=3 \\ \bf \beta=1$

Sum of zeroes:-

$\bf \alpha +\beta =3+1=\dfrac{4}{1} \\ \bf \to \dfrac{4×2}{1×2}=\dfrac{8}{2}=\dfrac{-b}{a}$

Product of zeroes:-

$\bf \alpha ×\beta =3×1=\dfrac{3}{1} \\ \bf \to \dfrac{3×2}{1×2}=\dfrac{6}{2}=\dfrac{c}{a}$