Question

The zeroes of the quadratic polynomial \[{{x}^{2}}+99x+127\] are:
A. both positive
B. both negative
C. one positive and one negative
D. both equal

Answer 1

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.

Complete step-by-step answer:
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.
In a quadratic polynomial, if,
\[\begin{align}
  & a>0 \\
 & a<0 \\
\end{align}\] or \[\left. \begin{matrix}
   b>0,c>0 \\
   b<0,c<0 \\
\end{matrix} \right\}\] then both zeroes are negative.
In the given polynomial, we see that a = 1 > 0, b = 99> 0 and c = 127 > 0.
Thus by this condition, both zeroes of the given quadratic polynomial are negative.