Question

How do you find the value of discriminant and state the type of solutions given $2{p^2} + 5p - 4 = 0$?

Answer 1

Hint: In this question, we are given an equation and we have to find its discriminant. In addition to this, we also have to state the type of solutions. For this, you must be aware about the type of solutions that different discriminants give and also the formula of discriminant. Then, find the discriminant using the formula and use the rules to find out the type of solutions.

Formula used: Discriminant D$ = {b^2} - 4ac$

Complete step-by-step solution:
We are given an equation $2{p^2} + 5p - 4 = 0$. This equation is a quadratic equation. Let us compare the given equation with the standard quadratic equation $a{x^2} + bx + c = 0$.
On comparing, we get,
$a = 2$,
$b = 5$, and
$c = - 4$.
We will put these values in the formula of discriminant.
$ \Rightarrow D = {b^2} - 4ac$
Putting the values,
$ \Rightarrow D = {5^2} - 4 \times 2 \times \left( { - 4} \right)$
On simplifying, we will get,
$ \Rightarrow D = 25 + 32$
$ \Rightarrow D = 57$
Now, since our discriminant is positive $(D > 0)$, we can say that our solutions are distinct and real.
In addition to this, let us find the solutions also.
The formula used to find the solutions is $ \to x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$.
We have all the values. Let us put them in the formula –
$ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {57} }}{4}$
Thus, these are our solutions.

Note: Let us know about the rules of discriminants:
1) If D is positive $(D > 0)$, the two solutions are distinct and real.
2) If D is zero $(D = 0)$, the two solutions are equal and real.
3) If D is negative $(D < 0)$, the two solutions are distinct and unreal.
Notice that we are using the terms “two solutions”. This is because a quadratic equation always has two solutions. The degree of the question indicates the number of solutions.