The given quadratic polynomial is p(x)=x2+99x+127
On comparing p(x) with ax2+bx+c, we get
a = 1, b = 99 and c = 127
We known that, x=−b±√b2−4ac2a
= −99±√(99)2−4×1×1272×1
= −99±√9801−5082
= −99±√92932=−99±9642
= −99+96.42,−99−96.42
= 2.62,−195.42
= −13,−977
Hence, both zeroes of the given quadratic polynomial p (x) are negative
Alternative Method
We know that,
In quadratic polynomial, if a>0a<0 or b>0,c>0b<0,C<0} , then both zeroes are negative
In given polynomial, we see that
a=1>0,b=99>and c=127>0
which satisfy the above condition
so, both zeroes of the given quadratic polynomial are negative