By analyzing the determinant, we will see that we will have two irrational solutions.
How to know the type of solutions of a quadratic equation?
To do it, we need to analyze the determinant.
For a general quadratic equation:
a*x^2 + b*x + c = 0
The determinant is:
D = b^2 - 4*a*c
- If D > 0, we have two real solutions.
- If D = 0, we have 1 real solution.
- if D < 0, we have two nonreal solutions.
In our case, we have:
3x^2 - x - 5 = 0
Then the determinant is:
(-1)^2 - 4*3*(-5) = 1 + 4*3*5 = 61
Then we have two real solutions, also remember that the solutions are given by:
[tex]x = \frac{-b \pm \sqrt{D} }{2a}[/tex]
Then we would have the square root of 61 in our solution, and 61 is a prime number, so it will give an irrational outcome, from this we can conclude that we will have two irrational solutions.
If you want to learn more about quadratic equations, you can read:
https://brainly.com/question/1214333
