Question

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

Answer 1

Hint: 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
$0.5={\displaystyle \frac{1}{2}}$ .
Now we put the values in the formula and we have:
$D=\sqrt{{(2a)}^2-4\times a\times {\displaystyle \frac{1}{2}}}<0$
On simplifying we have:
$\sqrt{4a^2-2a}<0$
By taking the common factor out, it gives:
$2a(2a-1)<0$
We will solve both the values separately i.e.
$2a<0\Rightarrow a>{\displaystyle \frac{0}{2}}$ .
It gives
 $a>0$ .
In the second value we have:
 $2a-1<0\Rightarrow 2a<1$
So we have
 $a<{\displaystyle \frac{1}{2}}$ .
We can say that the value of a lies between
 $\left(0,{\displaystyle \frac{1}{2}}\right)$
Or, it can be written as
$0<a<{\displaystyle \frac{1}{2}}$ .

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
 $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ .
If we have
$D=0$ , then the equation has a real root, which is given by
$x={\displaystyle \frac{-b}{2a}}$ .