Question 89372
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 {{{4*x^2-3*x+3=0}}} ( notice {{{a=4}}}, {{{b=-3}}}, and {{{c=3}}})


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




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




{{{x = (3 +- sqrt( 9-4*4*3 ))/(2*4)}}} Square -3 to get 9  (note: remember when you square -3, you must square the negative as well. This is because {{{-3^2=-3*-3=9}}}.)




{{{x = (3 +- sqrt( 9+-48 ))/(2*4)}}} Multiply {{{-4*3*4}}} to get {{{-48}}}




{{{x = (3 +- sqrt( -39 ))/(2*4)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (3 +- i*sqrt(39))/(2*4)}}} Simplify the square root (note: since we cannot take the square root of a negative value, we must factor {{{sqrt(-39)}}} to {{{i*sqrt(39)}}} to make the radicand positive. 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 = (3 +- i*sqrt(39))/(8)}}} Multiply 2 and 4 to get 8




After simplifying, the quadratic has roots of


{{{x=3/8 + sqrt(39)/8*i}}} or {{{x=3/8 - sqrt(39)/8*i}}}


Notice if we graph the quadratic {{{y=4*x^2-3*x+3}}}, we get


{{{ graph( 500, 500, -14.625, 15.375, -12.5625, 17.4375, 4*x^2-3*x+3) }}} graph of {{{y=4*x^2-3*x+3}}}


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