Question 337545


{{{4x^2-13x+3=0}}} Start with the given equation.



Notice that the quadratic {{{4x^2-13x+3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=4}}}, {{{B=-13}}}, 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 = (-(-13) +- sqrt( (-13)^2-4(4)(3) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=-13}}}, and {{{C=3}}}



{{{x = (13 +- sqrt( (-13)^2-4(4)(3) ))/(2(4))}}} Negate {{{-13}}} to get {{{13}}}. 



{{{x = (13 +- sqrt( 169-4(4)(3) ))/(2(4))}}} Square {{{-13}}} to get {{{169}}}. 



{{{x = (13 +- sqrt( 169-48 ))/(2(4))}}} Multiply {{{4(4)(3)}}} to get {{{48}}}



{{{x = (13 +- sqrt( 121 ))/(2(4))}}} Subtract {{{48}}} from {{{169}}} to get {{{121}}}



{{{x = (13 +- sqrt( 121 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (13 +- 11)/(8)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{x = (13 + 11)/(8)}}} or {{{x = (13 - 11)/(8)}}} Break up the expression. 



{{{x = (24)/(8)}}} or {{{x =  (2)/(8)}}} Combine like terms. 



{{{x = 3}}} or {{{x = 1/4}}} Simplify. 



So the solutions are {{{x = 3}}} or {{{x = 1/4}}} 

  

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