Question 9839
{{{12x^2 - 6 = x}}}
{{{12x^2 - x - 6 = 0}}}


now work out {{{(b^2 - 4ac)}}} where a = 12, b = -1 and c = -6


--> {{{(-1)^2 - (4)*(12)*(-6)}}}
--> 1 + 288
--> 289


if {{{b^2 - 4ac}}} is zero --> 1 rational root 
if {{{b^2 - 4ac}}} is +ve --> 2 roots
 --> if {{{sqrt(b^2 - 4ac)}}} is a natural number then roots are rational
 --> if {{{sqrt(b^2 - 4ac)}}} remains a square root, then roots are irrational


Now, {{{sqrt(289)}}} is 17 so we have rational roots.


Using the full Quadratic formula, we have:


{{{x = (1 + 7)/24}}} and {{{x = (1 - 7)/24}}}
so, x = 8/24 and x = -6/24
--> x = 1/3, x = -1/4


jon.