Answer:
Real and irrational root
Step-by-step explanation:
Since a quadratic equation
,
Has two real different roots,
If the value of discriminant ![D=b^2-4ac>0](https://tex.z-dn.net/?f=D%3Db%5E2-4ac%3E0)
Has two equal real roots,
If ![D=0](https://tex.z-dn.net/?f=D%3D0)
Has two imaginary roots,
If ![D](https://tex.z-dn.net/?f=D%3C0)
Here, the given equation,
![x^2-5x-4=0](https://tex.z-dn.net/?f=x%5E2-5x-4%3D0)
![D=(-5)^2-4\times 1\times -4=25+16= 41>0](https://tex.z-dn.net/?f=D%3D%28-5%29%5E2-4%5Ctimes%201%5Ctimes%20-4%3D25%2B16%3D%2041%3E0)
Hence, the equation has the two unequal real roots,
Now, √D is an irrational number ( Which can not be expressed in the form of p/q in which q≠0 )
Therefore, the given equation has real and irrational root,
Third option is correct.