Question 223419

{{{4x^2=x}}} Start with the given equation.



{{{4x^2-x=0}}} Subtract x from both sides.



Notice that the quadratic {{{4x^2-x}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=4}}}, {{{B=-1}}}, and {{{C=0}}}



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



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



{{{x = (-(-1) +- sqrt( (-1)^2-4(4)(0) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-1}}}, and {{{C=0}}}



{{{x = (1 +- sqrt( (-1)^2-4(4)(0) ))/(2(4))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(4)(0) ))/(2(4))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-0 ))/(2(4))}}} Multiply {{{4(4)(0)}}} to get {{{0}}}



{{{x = (1 +- sqrt( 1 ))/(2(4))}}} Subtract {{{0}}} from {{{1}}} to get {{{1}}}



{{{x = (1 +- sqrt( 1 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (1 +- 1)/(8)}}} Take the square root of {{{1}}} to get {{{1}}}. 



{{{x = (1 + 1)/(8)}}} or {{{x = (1 - 1)/(8)}}} Break up the expression. 



{{{x = (2)/(8)}}} or {{{x =  (0)/(8)}}} Combine like terms. 



{{{x = 1/4}}} or {{{x = 0}}} Simplify. 



So the solutions are {{{x = 1/4}}} or {{{x = 0}}}