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,
$$\begin{array}{rl}& x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\\ & x={\displaystyle \frac{-99\pm \sqrt{{\left(99\right)}^2-4\times 1\times 127}}{2\times 1}}={\displaystyle \frac{-99\pm \sqrt{9801-508}}{2}}={\displaystyle \frac{-99\pm \sqrt{9293}}{2}}={\displaystyle \frac{-99\pm 96.4}{2}}\end{array}$$
Thus x = $${\displaystyle \frac{-99-96.4}{2}}$$ and x = $${\displaystyle \frac{-99+96.4}{2}}$$.
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,
$$\begin{array}{rl}& a>0\\ & a<0\end{array}$$ or $$\begin{array}{c}b>0,c>0\\ b<0,c<0\end{array}\}$$ 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.