Question 625897


{{{4x^2+x-1=0}}} Start with the given equation.



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



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



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



{{{x = (-1 +- sqrt( 1+16 ))/(2(4))}}} Rewrite {{{sqrt(1--16)}}} as {{{sqrt(1+16)}}}



{{{x = (-1 +- sqrt( 17 ))/(2(4))}}} Add {{{1}}} to {{{16}}} to get {{{17}}}



{{{x = (-1 +- sqrt( 17 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-1+sqrt(17))/(8)}}} or {{{x = (-1-sqrt(17))/(8)}}} Break up the expression.  



So the solutions are {{{x = (-1+sqrt(17))/(8)}}} or {{{x = (-1-sqrt(17))/(8)}}} 

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