Question

The sign of the quadratic polynomial $a{x}^2+bx+c$ is always positive, if?
A) a is positive and $b^2-4ac⩽0$
B) a is positive and $b^2-4ac⩾0$
C) a is any real number and $b^2-4ac⩽0$
D) a is any real number and $b^2-4ac⩾0$

Answer 1

Hint: A quadratic polynomial is a parabola on a graph whose concavity depends on the sign of the coefficient of $x^2$. In order to find the condition for the sign of the quadratic polynomial $a{x}^2+bx+c$ is always positive, we will apply the following facts:
1) Quadratic polynomial $a{x}^2+bx+c$ is always positive when it doesn’t intersect the x-axis at any point, so have no real zeroes.
2) In order to get the quadratic polynomial $a{x}^2+bx+c$always positive, we need an upward parabola.

Using the above facts, we will find the corresponding results and required conditions.

Complete step by step solution: A quadratic polynomial is a polynomial in a variable (like x) with degree 2. When represented on the graph, a quadratic polynomial is a parabola.
We know that a parabola of the form $y=a{x}^2+bx+c$ is a vertical parabola.
Now, it can be an upward parabola or downward parabola.
A downward parabola is the one when y tends to negative infinity for x tending to both positive and negative infinity.
An upward parabola is the one when y tends to positive infinity for x tending to both positive and negative infinity.
Thus, in order to get the quadratic polynomial $a{x}^2+bx+c$ always positive, it must be an upward parabola.
Now we know that, for a quadratic polynomial$a{x}^2+bx+c$,
$a>0$ represents an upward parabola whereas $a<0$ represents a downward parabola.
Thus, in order to have the quadratic polynomial $a{x}^2+bx+c$ always positive or an upward parabola, we must have,
$a>0$ …(i)
Now, we know that
A quadratic polynomial is a polynomial in a variable (like x) with degree 2, thus, it will have 2 zeroes, real or imaginary.
Now, in order to have the quadratic polynomial $a{x}^2+bx+c$ always positive, it must always remain above the x-axis,
Or there must not be an x for which the quadratic polynomial $a{x}^2+bx+c$is zero or negative.
Thus, the quadratic polynomial $a{x}^2+bx+c$ has imaginary zeroes
Now, for the quadratic equation $a{x}^2+bx+c=0$
The condition for having no real roots is its discriminant must be less than zero.
Discriminant of a quadratic equation is defined as: $D=b^2-4ac$
Now, to get discriminant of a quadratic equation less than zero, $b^2-4ac<0$
Thus, $b^2-4ac<0$ …(ii)
From (i) and (ii), we get,
$a>0$ and $b^2-4ac<0$
Hence, most suitable option is option a $a>0$ and $b^2-4ac⩽0$

Note: Another approach to the question can be solving by the graph of the polynomial. For the sign of the quadratic polynomial $a{x}^2+bx+c$ to be always positive, we need an upward parabola, which we will get when $a>0$. So, it is our first condition.
Now, we need the parabola to never cut axis, so its vertex should be above the x-axis.
Now, vertex of a parabola $y=a{x}^2+bx+c$ is $\left({\displaystyle \frac{-b}{2a}},{\displaystyle \frac{4ac-b^2}{4a}}\right)$
So, y-coordinate of vertex must be greater than zero.
${\displaystyle \frac{4ac-b^2}{4a}}>0$
Now, $a>0$
$4ac-b^2>0$
$b^2-4ac<0$
Hence, the required conditions are, $a>0$ and $b^2-4ac<0$