Question

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficient x 2 - 2 x minus 8 ​

Answer 1

Given :-

x² - x - 8

To Find :-

Zeroes

Solution :-

x² - x - 8

x² + 2x - 4x - 8

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

(x - 4)(x + 2)

Either

x - 4 = 0

x = 4

And,

x + 2 = 0

x = -2

Let α = 4 and β = -2

On comparing the equation with ax² + bx + c

Sum = -b/a

⇒ 4 + (-2) = -(-2)/1

⇒ 4 - 2 = 2

⇒ 2

Product = c/a

⇒ 4 × (-2) = -8/1

⇒ -8 = -8

Hence,Verified

Answer 2

Answer:

First step : Finding zeroes

Given,

${x}^{2} - 2x - 8 \\ = {x}^{2} - 4x + 2x - 8 \\ = x(x - 4) + 2(x - 4) \\ = (x - 4)(x + 2) \\ \\ \bf \: for \: zeroes \: \: \: either \\ \: x - 4 = 0 \: \: \: \therefore \: \pink{x = 4 }\\ \bf \: or \\ x + 2 = 0 \: \: \: \: \therefore \: \pink{ x = - 2} \\ \\ \bf \: so \: zeroes \: \: are \: \: \green{ \boxed{x = 4 \: \: or \: - 2}}$

Second step : The relationship between the zeros and the coefficient.

$\sf \: if \: \alpha \: \: and \: \beta \: are \: the \: zeroes \: of \: the \: \\ \sf polynomial \: \: \: a {x}^{2} + bx + c \: \: \: \: then \: \\ \\ \\ \sf \: sum \: of \: zeroes \: \: \: \: \alpha + \beta = - \frac{b}{a} \: \: \: \: \: \: and \\ \\ \sf \: \: product \: of \: zeroes \: \: \alpha \beta = \frac{c}{a}$

Zeroes of the polynomial is 4 and - 2

So,

sum of zeroes is = 4+(-2) = 4-2 = 2

and

product of zeroes is = 4x(-2) = -8

Verify :

From the polynomial

$\alpha + \beta = - \frac{ - 2}{1} = 2 \\ \\ \alpha \beta = \frac{ - 8}{1} = - 8$

So verified.