Question

For what value of $a$ is the inequality $a{x^2} + 2ax + 0.5 > 0$ valid throughout the entire number axis?

Answer 1

In the above question we have to find the value of $a$ . For this we need to solve the above inequality. We can see that we have a quadratic equation. So we will solve this by using the formula of discriminant.
We know the formula states that if we have quadratic equation of the form
$a{x^2} + bx + c > 0$ , then we have $a > 0$ and $D < 0$ .
Here $a$ is the coefficient of ${x^2}$, and it has to be greater than zero, otherwise the quadratic equation will change.
The value of discriminant is
$D = \sqrt {{b^2} - 4ac}$ .

Complete answer: Here we have
$a{x^2} + 2ax + 0.5 > 0$ .
We know that if the quadratic equation is more than zero, i.e. it is a positive quadratic expression for all real value of $x$ . The graph will always be above x-axis and it will not have any real roots.
So the solutions to this equations will be imaginary and discriminant is always less than zero.
We can write it as
$a > 0$ and
$D < 0$ .
We know the formula
$D = \sqrt {{b^2} - 4ac}$ .
Here we have
$b = 2a,a = a$ and $c = 0.5$
Or, it can be written as
$0.5 = \dfrac{1}{2}$ .
Now we put the values in the formula and we have:
$D = \sqrt {{{(2a)}^2} - 4 \times a \times \dfrac{1}{2}} < 0$
On simplifying we have:
$\sqrt {4{a^2} - 2a} < 0$
By taking the common factor out, it gives:
$2a(2a - 1) < 0$
We will solve both the values separately i.e.
$2a < 0 \Rightarrow a > \dfrac{0}{2}$ .
It gives
$a > 0$ .
In the second value we have:
$2a - 1 < 0 \Rightarrow 2a < 1$
So we have
$a < \dfrac{1}{2}$ .
We can say that the value of a lies between
$\left( {0,\dfrac{1}{2}} \right)$
Or, it can be written as
$0 < a < \dfrac{1}{2}$ .

Note:
We should note that if the value of discriminant is greater than zero, i.e.
$D > 0$ , then the equation has two real and distinct roots.
They are given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .
If we have
$D = 0$ , then the equation has a real root, which is given by
$x = \dfrac{{ - b}}{{2a}}$ .