How do you solve #4x^2-8x+3=0#?
1 Answer
Jan 12, 2017
Explanation:
y = 4x^2 - 8x + 3 = 0
Use the new Transforming Method (Socratic Search)
Transformed equation:
y' = x^2 - 8x + 12.
Factor pairs of (ac = 12) --> (2, 6). This sum is (8 = -b). There for the 2 real roots of y' are: 2 and 6.
The 2 real roots of y are: x1 = 2/a = 2/ 4 = 1/2, and x2 = 6/a = 6/4 = 3/2
