Question

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

Answer 1

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 ${b^2} - 4ac \leqslant 0$

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 $\left( {\dfrac{{ - b}}{{2a}},\dfrac{{4ac - {b^2}}}{{4a}}} \right)$
So, y-coordinate of vertex must be greater than zero.
$\dfrac{{4ac - {b^2}}}{{4a}} > 0$
Now, $a > 0$
$4ac - {b^2} > 0$
${b^2} - 4ac < 0$
Hence, the required conditions are, $a > 0$ and ${b^2} - 4ac < 0$