Question

Question 7
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

The given quadratic polynomial is $p\left(x\right)=x^2+99x+127$
On comparing p(x) with $a{x}^2+bx+c$, we get
a = 1, b = 99 and c = 127
We known that, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$=\frac{-99\pm \sqrt{(99{)}^2-4\times 1\times 127}}{2\times 1}$
$=\frac{-99\pm \sqrt{9801-508}}{2}$
$=\frac{-99\pm \sqrt{9293}}{2}=\frac{-99\pm 964}{2}$
$=\frac{-99+96.4}{2},\frac{-99-96.4}{2}$
$=\frac{2.6}{2},\frac{-195.4}{2}$
$=-13,-977$
Hence, both zeroes of the given quadratic polynomial p (x) are negative
Alternative Method
We know that,
In quadratic polynomial, if $\begin{array}{c}\frac{a>0}{a<0}or\frac{b>0,c>0}{b<0,C<0}\end{array}\}$ , then both zeroes are negative
In given polynomial, we see that
$a=1>0,b=99>andc=127>0$
which satisfy the above condition
so, both zeroes of the given quadratic polynomial are negative