Question

Find the zeroes of the quadratic polynomial ${{x}^{2}}+99x+127$ and choose the correct option.
A. both positive
B. both negative
C. one positive and one negative
D. both equal

Answer 1

Hint: We first equate the given polynomial with the general form of quadratic equation. We try to find the points at where the curve intersects the X-axis. We take the x coordinates of those points using the theorem $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. At the end we evaluate the result with the given options.

Complete step-by-step answer:
We have a quadratic equation ${{x}^{2}}+99x+127$. Let $y\left( x \right)={{x}^{2}}+99x+127$.
We are finding the zeros of the polynomial. The solutions are the points of x at which the polynomial value is 0. In graphical form we are finding the intersection points of the curve with the X-axis.
Now we verify it with the algebraic version of the solution.
We use the theorem $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for the general equation of polynomial $a{{x}^{2}}+bx+c=0$.
So, at those root points the equational value is 0. So, we are solving the equation $y\left( x \right)={{x}^{2}}+99x+127$. Here $a=1,b=99,c=127$.
Putting values of $a=1,b=99,c=127$ in the equation $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
$x=\dfrac{-99\pm \sqrt{{{99}^{2}}-4\times 1\times 127}}{2\times 1}=\dfrac{-99\pm \sqrt{9293}}{2\times 1}$.
Now the value of $\sqrt{9293}$ is approximately 96.4. So, both values of $-99\pm \sqrt{9293}$ remain negative. They can’t be imaginary. They both are distinct values. So, roots can’t be equal.
So, the zeroes of the polynomial ${{x}^{2}}-3x+2$ are both negative. The correct option is B.

So, the correct answer is “Option B”.

Note: We need to understand that the polynomial value has to be 0. Zeroes of the polynomial are the roots of the polynomial. So, at those points the functional value of the curve is 0. The slope of the curve at those points is similar in value wise. We can also verify this result by substituting values of zeros in the given equation ${{x}^{2}}-3x+2$ and check if the results satisfy or not. The $\sqrt{{{b}^{2}}-4ac}$ part in the form of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the determinant.