Question

find the zero of the quadratic polynomial and verify the relationship between the zero and the coefficient x2-2x-8​

Answer 1

Answer:

Step-by-step explanation:

x^2-2x-8

= (x^2-4x)(+2x-8)

=x (x-4) +2 (x-4)

= (x-4) (x+2)

Therefore

(x-4) (x+2)=0

x=4 x=-2

Sum of zeroes = alpha+beeta= -b/a = 2/1=2

Product of zeroes= alpha×beeta= c/a= -8/1=-8

Therefore the zeroes are

2 and -8

Answer 2

$p(x) = {x}^{2} - 2x - 8$

To find the zeroes

p(x) = 0

$p(x) = {x}^{2} - 2x - 8 = 0 \\ \implies {x}^{2} + 2x - 4x - 8 =0 \\ \implies x(x + 2) -4(x+2) = 0 \\ \implies (x-4) (x+2) = 0$

x - 4 = 0

x = 4

x + 2 = 0

x = -2

Zeroes of ${x}^{2} - 2x - 8$ are -2 and 4

To verify the relationship

Sum of zeroes = -2 + 4 = 2

Product of zeroes = -2 × 4 = -8

Sum of zeroes = $\dfrac {- Coefficient \; of \; x}{Coefficient \; of \; {x}^{2}}$$= \dfrac {-(-2)}{1} = 2$

Product of zeroes = $\dfrac {Constant\; term}{Coefficient \; of \; {x}^{2}}$

$=\dfrac{-8}{1} =-8$

Hence, verified.