Question

Find the zeroes of the quadratic polynomial and verify the relationship between the zeros and the coefficient
=
$x {}^{2} - 2x - 8$

Answer 1

x²-2x-8
x²-4x+2x-8
x(x-4)+2(x-4)
(x+2)(x-4)=0
x=-2 , 4
sum of zeroes=-B/a
=-2

product=c/a
-8

Answer 2

${x}^{2} - 2x - 8 = 0 \\ {x}^{2} - 4x + 2x - 8 = 0 \\ x(x - 4) + 2(x - 4) = 0 \\ (x - 4)(x + 2) = 0 \\ (x - 4) = 0 \\ x = 4 \\ x + 2 = 0 \\ x = - 2 \\ we \: know \: that\\ \alpha + \beta = - \frac{b}{a} \\ = - \frac{ - 2}{1} = 2 \\ from \: zeros \: obtained \: sum \: of \: zeros = 4 + ( - 2) = 2 \\ now \: product \: will \: be \: (4)( - 2) = - 8 \\ \alpha \beta = \frac{c}{a} \\ \alpha \beta = \frac{ - 8}{1} = - 8 \\ hence \: proved$