Question

Please help :) Describe the nature of the roots for this equation. 3x^2+x-5=0

a. Two complex roots
b. Two real, rational roots
c. One real, double root
d. Two real, irrational roots

Answer 1

Answer: D

Step-by-step explanation: A-pex

Answer 2

`3x^2+x-5=0 \\ \\ a=3 \\ b=1 \\ c=-5 \\ \Delta=b^2-4ac=1^2-4 * 3 * (-5)=1+60=61`

The discriminant Δ is greater than 0, so there are two real roots.


`x=(-b \pm √(\Delta))/(2a)=(-1 \pm √(61))/(2 * 3)=(-1 \pm √(61))/(6)`

√61 is an irrational number, so both roots will be irrational.

The answer is D.