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

Comparing the given polynomial with `ax^(2)+bc+c`, we get
a=1, b=99, c=127
Now, `" " x=(-b+-sqrt(b^(2)-4ac))/(2a)`
`=(-99+-sqrt(99^(2)-4xx1xx127))/(2xx1)`
`=(-99+-sqrt(9801-508))/(2)=(-99+-sqrt(9293))/(2)`
`=(-99+-96.4)/(2)`
`=(-99+69.4)/(2)" " or " " (-99-96.4)/(2)`
`=-1.3-97.7`
`:.` Both zeroes are negative.
Alternative (Short cut) method:
Let two zeroes of given polynomial be `alpha` and `beta`.
`:.` Sum of zeroes `alpha+beta=-(b)/(a)=-(99)/(1)=-99`
and product of zeroes `alphabeta=(c )/(a)=(127)/(1)=127`
Since, product of zeroes is 127 i.e., positive, therefore both `alpha` and `beta` are either positive or both are negative. If we take both zeroes as positive, then their sum cannot be negative. But sum of zeroes is negative, so both the zeroes must be negative.