Question

19). Find the quadratic polynomial whose zeroes are 1 and -3, verify the relation between the coefficients & the zeroes of the polynomial.​

Answer 1

Zeroes : 1 and -3

==> p(x) = x² - ( sum of zeroes ) x + product of zeroes

==> p(x) = x² - ( 1 - 3 )x + (1 × -3)

==> p(x) = x² + 2x - 3

On comparing,

a = 1 ; b = 2 ; c = -3

Sum of Zeroes = -b/a

==> 1 + (-3) = -2/1

==> 1 - 3 = -2

==> - 2 = -2

Product of Zeroes = c/a

==> 1 × (-3) = -3/1

==> -3 = -3

Hence, Proved.

Answer 2

Answer:

x2+x-3

Step-by-step explanation:

let alpha=1 and bitta =-3

alpha +bitta =-coefficient of x÷coefficient of x2

1+(-3)=-coefficient of x÷coefficient of x2

1-3÷1=-coefficient of x ÷ coefficient of x2

-2÷1= -coefficient of x÷coefficient of x2

2÷1=coefficient of x ÷coefficient of x2

coefficient of x=2

coefficient of x2=1

product of alpha and bitta =constant term÷ coefficient of x2

1×-3=constant term ÷1

-3=constant term

let put the value

x2+2x-3