How do you solve #49x ^ { 2} - 56x + 15= 0#?

1 Answer
May 28, 2017

Answer:

#5/7 and 3/7#

Explanation:

Solve this quadratic equation methodically:
#y = 49x^2 - 56x + 15 = 0#
Use the new Transforming Method (Socratic, Google Search)
Transformed equation:
#y' = x^2 - 56x + 735 = 0# (ac = 49(15) = 735)
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:
#x1 = 35/a = 35/49 = 5/7# and #x2 = 21/(a) = 21/49 = 3/7#