How do you solve using the quadratic formula #3x(x - 2) = 1#?

Question

Kennedy Adams · Accepted Answer

f(x) = 3x^2 - 6x - 1 = 0 #D = d^2 = b^2 - 4ac = 36 + 12 = 48 = 16.3  -> 
d = +-4.sqr3# #x = 6/6 + (4.sqr3)/6#= #1 + (sqr3)/3# #x = 6/6 - (sqr3)/3 = 1 - (sqr3)/3#

Ethan Adkins · Answer

undefined To solve the equation $3x(x - 2) = 1$ using the quadratic formula, first, we need to rewrite the equation in standard quadratic form $ax^2 + bx + c = 0$. Then, we can identify $a$, $b$, and $c$ and substitute them into the quadratic formula:

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

Given equation: $3x(x - 2) = 1$

Expanding and rearranging the equation:

$3x^2 - 6x - 1 = 0$

Now, we can identify $a$, $b$, and $c$:

$a = 3$, $b = -6$, $c = -1$

Substituting into the quadratic formula:

$x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4 \cdot 3 \cdot (-1)}}}}{{2 \cdot 3}}$

$x = \frac{{6 \pm \sqrt{{36 + 12}}}}{{6}}$

$x = \frac{{6 \pm \sqrt{{48}}}}{{6}}$

$x = \frac{{6 \pm 4\sqrt{3}}}{{6}}$

Now, we can simplify further:

$x = \frac{{3 \pm 2\sqrt{3}}}{{3}}$

Therefore, the solutions to the equation $3x(x - 2) = 1$ using the quadratic formula are:

$x = \frac{{3 + 2\sqrt{3}}}{{3}}$ and $x = \frac{{3 - 2\sqrt{3}}}{{3}}$