Question

Determine the type and number of solutions of 4x^2-5x+1=0

a. two real solutions
b. one solution
c. tow imaginary solutions

Answer 1

4x^2-5x=-1

`√(4x^2-5x) = √(-1)`4x^2-5x+1=0

4x+2.24x=1

6.24x=1

x= 0.16025641025641025641025641025641

Answer 2

Two real solutions (x' = -1 and x'' = -1/4)