How do you solve #21n^2+114n=72#?

1 Answer
Jan 12, 2017

Answer:

4/7 and -6

Explanation:

After simplification by 3
f(n) = 7n^2 + 38n - 24 = 0.
Use the new Transforming Method (Socratic Search).
Transformed equation:
f'(n) = n^2 + 38n - 168 = (n - p')(n - q')
Find 2 numbers p' and q' that have opposite signs (ac < 0), knowing sum (-38) and product (-168).
Compose factor pairs of (ac = - 168) --> ...(4, -42). This last sum is
(- 38 = -b). There for, the 2 real roots of transformed equation are: p' = 4 and q' = - 42.
Back to original equation f(n), the 2 real roots are: p = (p')/a = 4/7,
and q = (q')/a = -42/7 = -6
NOTE.
1. This new method proceeds by finding the 2 real roots of the transformed equation. Then it divides these 2 real roots by (a) to get the 2 real roots of original equation.
2. This method avoids the lengthy factoring by grouping and solving the 2 binomials.