How do you solve $8 x^{2} - 6 x + 1 = 0$ $8x^2-6x+1=0$ ?

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Answer 1 · 2017-03-30T20:33:05

$x = \frac{1}{4} or x = \frac{1}{2}$ $x=1/4 or x=1/2$

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

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

(Find factors of 8 which add up 6.)

$(4 x - 1) (2 x - 1) = 0$ $(4x-1)(2x-1)=0$

If $4 x - 1 = 0 \to 4 x = 1 \to x = \frac{1}{4}$ $4x-1=0" " rarr 4x=1" "rarr x =1/4$

If $2 x - 1 = 0 \to 2 x = 1 \to x = \frac{1}{2}$ $2x-1=0" "rarr2x=1" "rarrx=1/2$

Answer 2 · 2017-03-31T21:05:20

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

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

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

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
$x = - (\frac{b}{2 \cdot a})$ $x = - ( b / (2*a))$

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

$x = \frac{- b \pm \sqrt{D}}{2 \cdot a}$ $x = ( -b +- sqrtD ) / ( 2 * a )$

Let's take your example:
$8 x^{2} - 6 x + 1 = 0$ $8x^2 - 6x + 1 = 0$

$D = 36 - 32 = 4 D > 0$ $D= 36 - 32 = 4 D>0$

$x 1 = \frac{6 + 2}{16}$ $x1= (6 + 2) / 16$
$x 1 = \frac{1}{2}$ $x1= 1 / 2$

And
$x 2 = \frac{6 - 2}{16}$ $x2= ( 6 - 2) / 16$
$x 2 = \frac{1}{4}$ $x2= 1 / 4$

How do you solve 8x^2-6x+1=08x2−6x+1=08x^2-6x+1=0?