Question

show the relationship between zeroes and coefficients of quadric equation x² - 2 x - 8 

Answer 1

x² - 2x - 8

x² - 4x + 2x - 8

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

( x - 4 ) ( x + 2 )


* ( x - 4 ) = 0
x = 4

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


=>
sum of zeros =
$\frac{ - coefficient \: of \: {x}}{coefficient \: of \: {x}^{2} }$

• LHS

Sum of Zeros :- 4 - 2 = 2

• RHS

$\frac{ - coeff \: of \: x}{coeff \: of \: {x}^{2} } = \frac{ 2}{1}$
LHS = RHS


=> Product of Zeros

$\frac{constant \: term}{coefficient \: of \: {x}^{2} }$

• LHS

Product of Zeros :- 4 × ( - 2 ) = - 8

• RHS

$\frac{constant \: term}{coefficient \: of \: {x}^{2} } = \frac{ - 8}{1}$
LHS = RHS

hence Zeros are verified by the coefficient of respective polynomial.

Answer 2

HEYA!!!!

Here is your answer :

${x}^{2} + 2x - 8 \\ \\ after \: splitting \: \\ \\ {x}^{2} + 2x - 4x - 8 \\ \\ x(x + 2) - 4(x + 2) \\ \\ x - 4 = 0 \: \: \: \: x + 2 = 0 \\ \\ x = 4 \: \: \: x = - 2 \\ \\ let \: \alpha = 4 \: \beta = - 2 \\ \\ using \: zeroes - \\ \\ \\ \alpha + \beta = 4 + ( - 2) \\ \\ \: \: \: \: \: \: \: = 4 - 2 = 2 \\ \\ \alpha \beta = 4 \times - 2 \\ \\ = - 8 \\ \\ using \: coefficients - \\ \\ \\ {x}^{2} - 2x - 8 \\ \\ a = 1 \: b \: = - 2 \: c = - 8 \\ \\ \alpha + \beta = \frac{ - b}{a} = 2 \\ \\ \alpha \beta = \frac{c}{a} = - 8$

HOPE THIS HELPS U. .