Question

diagram.
16.
Find the zeroes of the quadratic polynomial P(x) = x2 + x-12 and
verify the relationship between the zeroes and the coefficients.​

Answer 1

Answer:

x^2 +4x-3x-12=0

x(x+4)-3(x+4)=0

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

(x+4)=0. or (x-3)=0

x= -4 or x= +3

$\alpha = - 4 \: and \: \beta = 3$

On comparing the equation x^2+x-12=0 with ax^2 +bx +c =0 we get,

1.Sum of zeroes =

1.Sum of zeroes=

$\alpha + \beta = - 4 + 3 = - 1$

-coefficient of x/coefficient of x^2= -b/a=-1/1= -1

2.Product of zeroes=

$\alpha \times \beta = - 4 \times 3 = - 12$

c/a= -12/1= -12

$\alpha + \beta = - \frac{b}{a}$

$\alpha \times \beta = \frac{c}{a}$

Hence Verified...

Took more time to write answer sorry for delay...Hope it helps you...