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

May 1, 2017

shift everything to one side to give a quadratic equation

#### Explanation:

$3 {q}^{2} - 16 q + 5 = 0$
$\left(3 q - 1\right) \left(q - 5\right) = 0$
$q = 5$ and $q = \frac{1}{3}$

May 1, 2017

$q = \frac{1}{3} \mathmr{and} q = 5$

#### Explanation:

$3 {q}^{2} - 16 q = - 5$

$\therefore 3 {q}^{2} - 16 q + 5 = 0$

$\therefore = \left(3 q - 1\right) \left(q - 5\right)$

$\therefore 3 q = 1$

$\therefore \textcolor{p u r p \le}{q} = \textcolor{p u r p \le}{\frac{1}{3}}$ or color(purple)(q=5

substitute $q = \textcolor{p u r p \le}{\frac{1}{3}}$

$\therefore 3 {\left(\textcolor{p u r p \le}{\frac{1}{3}}\right)}^{2} - 16 \left(\textcolor{p u r p \le}{\frac{1}{3}}\right) = - 5$

$\therefore 3 \left(\frac{1}{9}\right) - \frac{16}{3} = - 5$

$\therefore \frac{3}{9} - \frac{16}{3} = - 5$

$\therefore \frac{3 - 48}{9} = - 5$

$\therefore - \frac{45}{9} = - 5$

$\therefore - 5 = - 5$

substitute color(purple)(q=5

$\therefore 3 {\left(\textcolor{p u r p \le}{5}\right)}^{2} - 16 \left(\textcolor{p u r p \le}{5}\right) = - 5$

$\therefore \left(3 \times 25\right) - 80 = - 5$

$\therefore 75 - 80 = - 5$

$\therefore - 5 = - 5$

May 24, 2017

$\frac{1}{3}$ and 5

#### Explanation:

$y = 3 {q}^{2} - 16 q + 5 = 0$
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
$y = 3 {q}^{2} - q - 15 q + 5 = q \left(3 q - 1\right) - 5 \left(3 q - 1\right) = \left(3 q - 1\right) \left(q - 5\right)$
(3q - 1) = 0 --> 3q = 1 --> $q = \frac{1}{3}$
q - 5 = 0 --> q = 5
2. The New Transforming Method (Socratic Search)
$y = 3 {q}^{2} - 16 q + 5 = 0$
Transformed equation:
$y ' = {q}^{2} - 16 q + 15 = 0$
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: $x 1 = \frac{1}{a} = \frac{1}{3} ,$ and $x 2 = \frac{5}{a} = \frac{5}{3}$