Question 227951


{{{4x^2-4x+3=0}}} Start with the given equation.



Notice that the quadratic {{{4x^2-4x+3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=4}}}, {{{B=-4}}}, and {{{C=3}}}



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



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-4) +- sqrt( (-4)^2-4(4)(3) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-4}}}, and {{{C=3}}}



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



{{{x = (4 +- sqrt( 16-4(4)(3) ))/(2(4))}}} Square {{{-4}}} to get {{{16}}}. 



{{{x = (4 +- sqrt( 16-48 ))/(2(4))}}} Multiply {{{4(4)(3)}}} to get {{{48}}}



{{{x = (4 +- sqrt( -32 ))/(2(4))}}} Subtract {{{48}}} from {{{16}}} to get {{{-32}}}



{{{x = (4 +- sqrt( -32 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (4 +- 4i*sqrt(2))/(8)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (4+4i*sqrt(2))/(8)}}} or {{{x = (4-4i*sqrt(2))/(8)}}} Break up the expression.  



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



So the solutions are {{{x = (1+i*sqrt(2))/(2)}}} or {{{x = (1-i*sqrt(2))/(2)}}} 



Note: if you have no idea what the 'i' terms are all about, then you probably haven't learned about imaginary numbers yet. If that's the case, then there are no solutions to this problem.