Question 140894
{{{x^2-(4/3)x+1/3=0}}} Start with the given equation



{{{3(x^2-(4/cross(3))x+1/cross(3))=3*0}}} Multiply both sides by the LCD 3. This will cancel out the denominators and clear the fractions



{{{3x^2-4x+1=0}}} Distribute and multiply



Now 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^2-4*x+1=0}}} ( notice {{{a=3}}}, {{{b=-4}}}, and {{{c=1}}})





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




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




{{{x = (4 +- sqrt( 16-4*3*1 ))/(2*3)}}} Square -4 to get 16  (note: remember when you square -4, you must square the negative as well. This is because {{{(-4)^2=-4*-4=16}}}.)




{{{x = (4 +- sqrt( 16+-12 ))/(2*3)}}} Multiply {{{-4*1*3}}} to get {{{-12}}}




{{{x = (4 +- sqrt( 4 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (4 +- 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 = (4 +- 2)/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{x = (4 + 2)/6}}} or {{{x = (4 - 2)/6}}}


Lets look at the first part:


{{{x=(4 + 2)/6}}}


{{{x=6/6}}} Add the terms in the numerator

{{{x=1}}} Divide


So one answer is

{{{x=1}}}




Now lets look at the second part:


{{{x=(4 - 2)/6}}}


{{{x=2/6}}} Subtract the terms in the numerator

{{{x=1/3}}} Divide


So another answer is

{{{x=1/3}}}


So our solutions are:

{{{x=1}}} or {{{x=1/3}}}


Notice when we graph {{{3*x^2-4*x+1}}}, we get:


{{{ graph( 500, 500, -9, 11, -9, 11,3*x^2+-4*x+1) }}}


and we can see that the roots are {{{x=1}}} and {{{x=1/3}}}. This verifies our answer