Question

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

Answer 1

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`

Answer 2

`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`

Answer 3

`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`