# How do you solve 8x^2-6x+1=0?

Mar 30, 2017

$x = \frac{1}{4} \mathmr{and} x = \frac{1}{2}$

#### Explanation:

Factorise the quadratic trinomial and then let each factor be equal to 0.

$8 {x}^{2} - 6 x + 1 = 0$

(Find factors of 8 which add up 6.)

$\left(4 x - 1\right) \left(2 x - 1\right) = 0$

If $4 x - 1 = 0 \text{ " rarr 4x=1" } \rightarrow x = \frac{1}{4}$

If $2 x - 1 = 0 \text{ "rarr2x=1" } \rightarrow x = \frac{1}{2}$

Mar 31, 2017

$x = \frac{1}{2}$ or $x = \frac{1}{4}$ (You usually solve equations of this type using the discriminant)

#### Explanation:

In any equation with the form of: $a {x}^{2} + b x + c = 0$ you can calculate the discriminant ( D or Δ) by using the formula:

$D = {b}^{2} - 4 \cdot a c$

Then, there are three different cases of finding the roots depending on the discriminant

No1: if D < 0 then the equation has no rootes

No2: if D = 0 then the equation has one answer which you can find using
$x = - \left(\frac{b}{2 \cdot a}\right)$

No3: if D>0 then the equation has two roots which you can find using

$x = \frac{- b \pm \sqrt{D}}{2 \cdot a}$

$8 {x}^{2} - 6 x + 1 = 0$
D= 36 - 32 = 4 D>0
$x 1 = \frac{6 + 2}{16}$
$x 1 = \frac{1}{2}$
$x 2 = \frac{6 - 2}{16}$
$x 2 = \frac{1}{4}$