QUESTION

The zeroes of the quadratic polynomial ${{x}^{2}}+99x+127$ are:A. both positiveB. both negativeC. one positive and one negativeD. both equal

Hint: The given quadratic equation is similar to the general equal of the quadratic equation $a{{x}^{2}}+bx+c=0$. Thus get the values of a, b and c. Substitute the values in quadratic formula and get the zeroes.

We have been given the quadratic polynomial ${{x}^{2}}+99x+127$.
Let us compare this quadratic polynomial with the general quadratic equation $a{{x}^{2}}+bx+c=0$, i.e. by comparing both ${{x}^{2}}+99x+127$ and the general equation $a{{x}^{2}}+bx+c=0$, we get,
a = 1, b = 99, c = 127.
Now let us substitute these values in the quadratic formula,
\begin{align} & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ & x=\dfrac{-99\pm \sqrt{{{\left( 99 \right)}^{2}}-4\times 1\times 127}}{2\times 1}=\dfrac{-99\pm \sqrt{9801-508}}{2}=\dfrac{-99\pm \sqrt{9293}}{2}=\dfrac{-99\pm 96.4}{2} \\ \end{align}
Thus x = $\dfrac{-99-96.4}{2}$ and x = $\dfrac{-99+96.4}{2}$.
x = -1.3 and x = -97.7.
Hence we got both the zeroes as negative, i.e. x = -1.3, -97.7.
Thus the given quadratic polynomial, ${{x}^{2}}+99x+127$ has both zeroes negative.
Option B is the correct answer.

Note: We can also find the zeroes of quadratic polynomials with an alternate method.
\begin{align} & a>0 \\ & a<0 \\ \end{align} or $\left. \begin{matrix} b>0,c>0 \\ b<0,c<0 \\ \end{matrix} \right\}$ then both zeroes are negative.