# How do you solve 3x² – 16x + 5 = 0?

Jun 28, 2016

$x = 5$ or $x = \frac{1}{3}$

#### Explanation:

Factorise into $\left(3 x - 1\right) \left(x - 5\right) = 0$

For the left hand side to be zero, the contents of one of the brackets must evaluate to zero, hence

$3 x - 1 = 0 \implies x = \frac{1}{3}$

or

$x - 5 = 0 \implies x = 5$.

Alternatively you can use the quadratic formula for equations of the from $a {x}^{2} + b x + c = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

So in this case, a = 3, b = -16 and c = 5.

$x = \frac{16 + \sqrt{256 - 60}}{6}$ or $\frac{16 - \sqrt{256 - 60}}{6}$

$\therefore x = 5 \mathmr{and} x = \frac{1}{3}$

Jun 29, 2016

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

#### Explanation:

Another way is to use the Transforming Method (Google, Yahoo Search)
$y = 3 {x}^{2} - 16 x + 5$.
Find the real roots of the transformed equation:
$y ' = {x}^{2} - 16 x + 15$.
Since a + b + c = 0, use shortcut, the 2 real roots are: X1 = 1 and
$X 2 = \frac{c}{a} = 15$.
Back to original equation y, the 2 real roots are: $x 1 = \frac{X 1}{a} = \frac{1}{3}$
and $x 2 = \frac{X 2}{a} = \frac{15}{3} = 5.$