Question 169892

{{{4x^2=2x+1}}} Start with the given equation.



{{{4x^2-2x-1=0}}} Get all terms to the left side.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=-2}}}, and {{{c=-1}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



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



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



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



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



{{{x = (2 +- sqrt( 4+16 ))/(2(4))}}} Rewrite {{{sqrt(4--16)}}} as {{{sqrt(4+16)}}}



{{{x = (2 +- sqrt( 20 ))/(2(4))}}} Add {{{4}}} to {{{16}}} to get {{{20}}}



{{{x = (2 +- sqrt( 20 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (2 +- 2*sqrt(5))/(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 = (2+2*sqrt(5))/(8)}}} or {{{x = (2-2*sqrt(5))/(8)}}} Break up the expression.  



{{{x = (1+sqrt(5))/(4)}}} or {{{x = (1-sqrt(5))/(4)}}} Reduce



So the answers are {{{x = (1+sqrt(5))/(4)}}} or {{{x = (1-sqrt(5))/(4)}}}  



which approximate to {{{x=0.809}}} or {{{x=-0.309}}}