# How do you solve 5x ^ { 2} - 17x + 6= 0?

Jun 14, 2017

Break the trinomial down into two binomials and solve for each binomial

#### Explanation:

$A {x}^{2} + B x + C$ is the standard form of the trinomial.

The B in this equation is negative.
The C in this equation is positive.

This means that for B to be negative both factors of C must be negative.

The sum of the products of the A factors times the C factors, must equal -17.

$\left(5 \times 3 = 15\right) + \left(1 \times 2 = 2\right)$

$- 15 + - 2 = - 17$ so

$\left(5 x - 2\right) \times \left(1 x - 3\right) = 5 {x}^{2} - 17 x + 6$ so

$5 x - 2 = 0$ add 2 to both sides

$5 x - 2 + 2 = 0 + 2$ this gives

$5 x = 2$ divide both sides by 5

$5 \frac{x}{5} = \frac{2}{5}$ this gives.

$x = \frac{2}{5}$

$1 x - 3 = 0$ add three to both sides.

$1 x - 3 + 3 = 0 + 3$ this gives.

$1 x = 3$

So x equals both 2/5 and 3

Jun 30, 2017

2/5 and 5

#### Explanation:

$y = 5 {x}^{2} - 17 x + 6 = 0$
Use the new Transforming Method (Socratic, Google Search)
Transformed equation:
$y ' = {x}^{2} - 17 x + 30 = 0$
Method: find 2 real roots of y', then, divide them by a = 5.
Find 2 real roots of y', knowing sum (-b = 17) and product (ac = 30).
They are : 2 and 15.
Back to original y, the 2 real roots are:
$x 1 = \frac{2}{a} = \frac{2}{5}$, and
$x 2 = \frac{15}{a} = \frac{15}{5} = 3$