Explanation:
Both negative
Let given quadratic polynomial be p(x) = x\(^{1}\) + 99x + 127. On comparing p(x) with ax\(^{1}\) + bx + c, we get a = 1, b = 99 and c = 127 We know that, \(\text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}\) [by quadratic formula] \(=\frac{-99\pm\sqrt{(99)^2-4\times1\times127}}{2\times1}\) \(=\frac{-99\pm\sqrt{9801-508}}{2}\) \(=\frac{-99\pm\sqrt{9293}}{2}=\frac{-99\pm96.4}{2}\) \(=\frac{-99+96.4}{2},\frac{-99-96.4}{2}\) \(=\frac{-2.6}{2},\frac{-195.4}{2}\) \(=-13, -97.7\) Hence, both zeroes of the given quadratic polynomial p(x) are negative. Alternate answer In quadratic polynomial, if \(\begin{matrix}\text{a} > 0 \\ \text{a} < 0 \end{matrix}\text{ or } \begin{matrix} \text{b} > 0,\text{ c} > 0 \\ \text{b}< 0, \text{ c} < 0 \end{matrix}\Bigg\},\) 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.
