Question

Find the zeroes of the following quadratic polynomials and verily the relationship between
the zeroes and the coefficients.
x2 - 2x -8​

Answer 1

Step-by-step explanation:

firstly by splitting middle term

x^2-4x+2x-8 = 0

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

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

x-4=0 x+2=0

x=4,x=-2 these are the 2 zeroes (alpha and beta)

*verification

alpha + beta = -b/a

4+(-2) = -(-2)/1

2 = 2

alpha*beta = c/a

4*(-2) = -8/1

8=8

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Answer 2

Step-by-step explanation:

take any 2 no. whose we have required no. -4 ,2

$\alpha = - 2 \\ \beta = - 8$

x

$x {}^{2} - 2x - 8$

$x {}^{2} - 4x + 2x - 8 x \\ x(x - 4) + 2 (x - 4) \\ (x - 4)(x + 2)$

zeroes of polynomial equation x^2-2x-8 are (-4,2)

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

x-4=0 ;x=4

x+2=0;x=-2

relation

now sum of the coeff. and zero

$\alpha = 4 + ( - 2) = \frac{ - \beta }{\alpha } \\ 4 - 2 = ( \frac{ - 2}{1} ) \\ 2 = 2$

$\beta = 4 \times ( - 2) = \frac{c}{ \alpha } \\ \\ - 8 = \frac{ - 8}{1?}$

Hence verified ..