How do you solve #49x ^ { 2} - 56x + 15= 0#?
1 Answer
May 28, 2017
Explanation:
Solve this quadratic equation methodically:
Use the new Transforming Method (Socratic, Google Search)
Transformed equation:
Proceeding: Find the 2 real roots of y', then, divide them by a = 49.
Find 2 real roots knowing sum (-b = 56) and product (ac = 735).
Compose factor pairs of (735) --> ...(5, 147)(7, 105)(21, 35).
This sum is (35 + 21 = 56 = -b). Therefore, the 2 real roots of y' are:
35, and 21. Back to original y, the 2 real roots are:
