How do you solve #3q^2 - 16q = -5#?

3 Answers
May 1, 2017

Answer:

shift everything to one side to give a quadratic equation

Explanation:

#3q^2 - 16q + 5 = 0#
#(3q-1)(q-5)=0#
# q = 5# and # q = 1/3#

May 1, 2017

Answer:

#q=1/3orq=5#

Explanation:

#3q^2-16q=-5#

#:.3q^2-16q+5=0#

#:.=(3q-1)(q-5)#

#:.3q=1#

#:. color(purple)q=color(purple)(1/3)# or #color(purple)(q=5#

substitute #q=color(purple)(1/3)#

#:.3(color(purple)(1/3))^2-16(color(purple)(1/3))=-5#

#:.3(1/9)-16/3=-5#

#:.3/9-16/3=-5#

#:.(3-48)/9=-5#

#:.-45/9=-5#

#:.-5=-5#

substitute #color(purple)(q=5#

#:.3(color(purple)5)^2-16(color(purple)5)=-5#

#:.(3 xx 25)-80=-5#

#:.75-80=-5#

#:.-5=-5#

May 24, 2017

Answer:

#1/3# and 5

Explanation:

#y = 3q^2 - 16q + 5 = 0#
There are 2 methods to choose:

  1. The AC Method --> split the middle term for factoring:
    Find 2 numbers knowing sum (b = -16) and product (ac = 15).
    They are -1 and - 15.
    Split - 16q into -q and - 15q
    #y = 3q^2 - q - 15q + 5 = q(3q - 1) - 5(3q - 1) = (3q - 1)(q - 5)#
    (3q - 1) = 0 --> 3q = 1 --> #q = 1/3#
    q - 5 = 0 --> q = 5
  2. The New Transforming Method (Socratic Search)
    #y = 3q^2 - 16q + 5 = 0#
    Transformed equation:
    #y' = q^2 - 16q + 15 = 0#
    Method: Find the 2 real roots of y', then divide them by a = 3
    Find 2 numbers (real roots) knowing sum (-b = 16) and product (ac = 15). They are: 1 and 15.
    Back to y, the 2 real roots are: #x1 = 1/a = 1/3,# and #x 2 = 5/a = 5/3#