The zeroes of the quadratic polynomial `x^(2)+99x +127` are

Options

A. both positive
B. both negative
C. one positive and one negative
D. both equal

Answer

B

Explanation

Let given quadratic polynomial be `p(x)=x^(2)+99x +127`.
On comparing `p(x)` with `ax^(2)+bx +c`, we get
`a = 1, b = 99` and `c = 127`
We know that, `x = (-b +- sqrt(b^(2)-4ac))/(2a)` [by quadratic formula]
`=(-99+-sqrt((99)^(2)-4xx1xx127))/(2xx1)`
`= (-99 +- sqrt(9801-508))/(2)`
`= (-99 +-sqrt(9293))/(2)=(-99+-96.4)/(2)`
`= (-99+96.4)/(2),(-99-96.4)/(2)`
`=(-2.6)/(2),(-195.4)/(2)`
`=- 1.3, 97.7`
Hence, both zeroes of the given quadratic polynomial `p(x)` are negative,
Alternate Method
In quadratic polynomilal, if `{:(agt 0,bgt 0 " "c gt 0),(" or",),(alt 0,b lt0 " "clt0):}}`, then both zeroes are negative. In given polynomial, we see that
`a = 1 gt 0, b = 00 gt 0` and `c = 127 gt 0`
which satisfy the above condition.
So, both zeroes of the given quadratic polynomial are negative.