Question

How do you find the value of the discriminant and determine the nature of the roots $ - 4{r^2} - 4r = 6$?

Answer 1

Hint:First of all we will make the above expression in the form of $a{x^2} + bx + c$ and then we will able to find the discriminant by using the formula $\vartriangle = {b^2} - 4ac$ and after solving we will identify the nature of the roots.

Formula used:
The expression for the quadratic equation is given by,
$a{x^2} + bx + c$
Here, $x$ will be constant
$a,b,c$ will be the variables.
Discriminant,
$\vartriangle = {b^2} - 4ac$
$\vartriangle $ , will be equal to discriminant.

Complete step by step answer:
So, we have the equation given as $ - 4{r^2} - 4r = 6$ .
Now taking all the terms at one side, and rearrange the equation in order, we will get the equation as
$ \Rightarrow - 4{r^2} - 4r - 6 = 0$
Taking the negative sign common we will get the equation as
$ \Rightarrow 4{r^2} + 4r + 6 = 0$
As the above equation is in the form of $a{x^2} + bx + c$ , so now we will calculate the discriminant from it.
For this the values will be as: $a = 4,b = 4,c = 6$
On substituting the above values in the expression of discriminant, we will get the equation as
$ \Rightarrow \vartriangle = {4^2} - 4 \times 4 \times 6$
Now on solving the squares and multiplication, we get
$ \Rightarrow \vartriangle = 16 - 96$
And on solving it we will get the equation as
$ \Rightarrow \vartriangle = - 80$
Therefore, on solving the value of the discriminant, we get $ - 80$ .
As we know if the value of $\vartriangle > 0$ , then there are two distinct roots and also having separate real roots.
Similarly, if the value of $\vartriangle = 0$ , then there is only one real root.
And if the value of $\vartriangle < 0$ , then there are no real roots.
So in this case we have a value $\vartriangle = - 80$ that is less than zero. Therefore, it will follow $\vartriangle < 0$

Hence, the nature of the expression has no real roots.

Note: Most of the time we are familiar with seeing the value $x$ as constant. And that’s why it becomes important for us to keep in mind that the constant can be anything we should get accustomed to by seeing the equation like this. In this problem, it will be $r$.