Question

How do you find the value of the discriminant and state the type of solutions given ${r^2} + 5r + 2 = 0$?

Answer 1

Hint: In this question we have to find the value of the discriminant and state the type of zeros, we will do this by using discriminant formula and ${b^2} - 4ac$ is called the discriminant, and the nature of the roots can be determined by using discriminant if the discriminant is negative then the roots of the quadratic equation will be complex numbers, and if the discriminant is positive then the roots of the quadratic equation will be real numbers, and if the discriminant is zero then the roots of the quadratic equation will be have one root, substituting the values in the discriminant we will get the required result.

Complete step by step answer:
Now the given quadratic equation is,
${r^2} + 5r + 2 = 0$,
As the equation has a degree 2 we will have two roots for the equation.
Now we will determine the discriminant using the formula which is given by ${b^2} - 4ac$, and the nature roots of any equation can be determined by using discriminant i.e., if the discriminant i.e., ${b^2} - 4ac$ is greater than zero then the equation will have real roots, if the discriminant i.e., ${b^2} - 4ac$ is equal to zero then the equation will have one real repeated root, if the discriminant i.e., ${b^2} - 4ac$ is less than zero then the equation will have complex roots.
So, here$a = 1$,$b = 5$,$c = 2$,
Now substituting the values in the discriminant i.e., ${b^2} - 4ac$ we get,
$ \Rightarrow {\left( 5 \right)^2} - 4\left( 1 \right)\left( 2 \right)$,
Now simplifying we get,
$ \Rightarrow 25 - 8 = 17$,
So,${b^2} - 4ac = 17 > 0$, according to the nature of the roots, the given equation will have two real roots.
Now using the quadratic formula, here it is given by $r = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$,
Here$a = 1$,$b = 5$,$c = 2$,
Now substituting the values in the formula we get,
$ \Rightarrow r = \dfrac{{ - \left( 5 \right) \pm \sqrt {{{\left( 5 \right)}^2} - 4\left( 1 \right)\left( 2 \right)} }}{{2\left( 1 \right)}}$,
Now simplifying we get,
$ \Rightarrow r = \dfrac{{ - 5 \pm \sqrt {25 - \left( 8 \right)} }}{2}$,
Now again simplifying we get,
$ \Rightarrow r = \dfrac{{ - 5 \pm \sqrt {17} }}{2}$,
Now we get two values of $r$they are $r = \dfrac{{ - 5 + \sqrt {17} }}{2}$and
,$r = \dfrac{{ - 5 - \sqrt {17} }}{2}$
So the zeros of the equation are $\dfrac{{ - 5 \pm \sqrt {17} }}{2}$, as the discriminant is greater than zero the given equation has two real roots.

$\therefore $The discriminant for the given equation, i.e., ${r^2} + 5r + 2 = 0$, is $\sqrt {17} $and the zeros of the equation will be $\dfrac{{ - 5 \pm \sqrt {17} }}{2}$, and the equation will have two real roots.

Note: The discriminant is part of the quadratic formula which lies underneath the square root. The quadratic equation discriminant is important because it tells us the number and type of solutions. This information is helpful because it serves as a double check when solving quadratic equations by any of the four methods i.e., factoring, completing the square, using square roots, and using the quadratic formula. If you are trying to determine the "type" of roots of any equation we need not complete the entire quadratic formula. Simply look at the discriminant.