How do you solve #20x ^ { 2} = 7x + 3#?

1 Answer
Nghi N
Jun 25, 2017

(4x + 1)(5x - 3)

Explanation:

#y = 20x^2 - 7x - 3 = 0 =# a(x + p)(x + q)
Use the new AC Method to factor trinomials (Google, Socratic Search)
Converted trinomial:
#y' = x^2 - 7x - 60 =# (x + p')(x + q')
Find to numbers knowing sum (b = -7) and product (ac = -60).
They are: p' = 5 and q' = - 12
Back to y, #p = (p')/a = 5/20 = 1/4#, and #q = (q')/20 = -12/20 = -3/5#
Factored form:
#y = 20(x + 1/4)(x - 3/5) = (4x + 1)(5x - 3)#

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