The zeroes of the quadratic polynomial

Solution:

(b) Let given quadratic polynomial be p(x) =x2 + 99x + 127.

On comparing p(x) with ax2 + bx + c, we get

a = 1, b = 99 and c = 127

We know that,   $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$   [by quadratic formula]

$=\frac{-99 \pm \sqrt{(99)^{2}-4 \times 1 \times 127}}{2 \times 1}$

$=\frac{-99 \pm \sqrt{9801-508}}{2}$

$=\frac{-99 \pm \sqrt{9293}}{2}=\frac{-99 \pm 96.4}{2}$

$=\frac{-99+96.4}{2}, \frac{-99-96.4}{2}$

$=\frac{-2.6}{2}, \frac{-195.4}{2}$

$=-1.3,-97.7$

Hence, both zeroes of the given quadratic polynomial p(x) are negative.
Alternate Method

In quadratic polynomial, if $\left.\begin{array}{ll}a>0 & \text { or } b>0, c>0 \\ a<0 & b<0, c<0\end{array}\right\}$, then both zeroes are negative

In given polynomial, we see that $a=1>0, b=99>0$ and $c=127>0$

the above condition.

So, both zeroes of the given quadratic polynomial are negative.