Question

Find the zeros of the following quadratic polynomials and verify the basic relationships
between the zeros and the coefficients.

Answer 1

HELLO DEAR,

let the roots of the equation be a and b.

( 1 ). x² - 2x - 8 = 0

x² - 4x + 2x - 8 = 0

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

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

a = x = -2 , b = x = 4

Relation,

a + b = -(cofficient of x)/(cofficient of x²)

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

2 = 2 [ verify]

a * b = (cofficient of constant term)/(cofficient of x²)

(-2) * (4) = (-8)/(1)

-8 = -8 [ verify]


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Let p(x) = x² - 2x - 8

We get zeroes of the polynomial

p(x) we will take p(x)=0

=> x² - 2x - 8 = 0

=> x² -4x + 2x - 8 = 0

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

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

x - 4 = 0 or x + 2 = 0

x = 4 or x = -2

The zeroes of x² - 2x - 8 are -2 , 4

i) Sum of zeroes = -2 + 4 = 2

= - (coeffient of x)/(Coefficient of x²)

ii ) Product of zeroes

= (-2) × 4

= -8

= (Constant term)/(Coefficient of x²)

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