How do you solve # 5x^2 - 12x + 10 = x^2 + 10x#?

1 Answer
Mar 11, 2016

Answer:

4 and 5

Explanation:

#y = 5x^2 - 12x + 10 - x^2 - 10x = 0#
#y = 4x^2 - 22x + 10 = 0.#
Use the new Transforming Method (Google, Yahoo).
Transformed equation: #y' = x^2 - 22x + 40 = 0#
Factor pairs of (ac = 40) --> (2, 20). This sum is 22 = -b.
Therefor, the 2 real roots of y' are: y1 = 2 and y2 = 20.
Back to original equation y, the 2 real roots are:
#x1 = (y1)/a = 2/4 = 1/2# and #x2 = (y2)/a = 20/4 = 5#
Answers: #1/2# and 5.
NOTE.
There is no need to factor by grouping and solving the 2 binomials.