Question

Find the zeroes of the polynomials x^2 + 6 root 6x + 48 and verify the relationship between zeroes and the coefficients.

Answer 1

Answer:

Factorize the equation, we get (x+2)(x−4)

So, the value of x

2

−2x−8 is zero when x+2=0,x−4=0, i.e., when x=−2 or x=4.

Therefore, the zeros of x

2

−2x−8 are -2 and 4.

Now,

⇒Sum of zeroes =−2+4=2=−

1

2

=−

Coefficient of x

2

Coefficient of x

⇒Product of zeros =(−2)×(4)=−8 =

1

−8

=

Coefficient of x

2

Constant term

Answer 2

Note: please consider # as root symbol
This can be factorized as (x+2#6)(x+4#6)=0
Hence two zeroes are x= -2#6 and x= -4#6

Adding two zeroes, -2#6 + (-4#6)= -6#6 = -(coefficient of x/coefficient of x^2)

Product of zeroes, -2#6 x (-4#6)= 48 = constant/coefficient of x^2

Verified