Question 93575
Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{3*x^2-2*x+1=0}}} ( notice {{{a=3}}}, {{{b=-2}}}, and {{{c=1}}})


{{{x = (--2 +- sqrt( (-2)^2-4*3*1 ))/(2*3)}}} Plug in a=3, b=-2, and c=1




{{{x = (2 +- sqrt( (-2)^2-4*3*1 ))/(2*3)}}} Negate -2 to get 2




{{{x = (2 +- sqrt( 4-4*3*1 ))/(2*3)}}} Square -2 to get 4  (note: remember when you square -2, you must square the negative as well. This is because {{{(-2)^2=-2*-2=4}}}.)




{{{x = (2 +- sqrt( 4+-12 ))/(2*3)}}} Multiply {{{-4*1*3}}} to get {{{-12}}}




{{{x = (2 +- sqrt( -8 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (2 +- 2*i*sqrt(2))/(2*3)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (2 +- 2*i*sqrt(2))/(6)}}} Multiply 2 and 3 to get 6




After simplifying, the quadratic has roots of


{{{x=1/3 + sqrt(2)/3*i}}} or {{{x=1/3 - sqrt(2)/3*i}}}


Notice if we graph the quadratic {{{y=3*x^2-2*x+1}}}, we get


{{{ graph( 500, 500, -14.6666666666667, 15.3333333333333, -14.3333333333333, 15.6666666666667, 3*x^2-2*x+1) }}} graph of {{{y=3*x^2-2*x+1}}}


And we can see that there are no real roots


To visually verify the answer, check out <a href=http://www.math.hmc.edu/funfacts/ffiles/10005.1.shtml>this page</a> to see a visual representation of imaginary roots