Question 342743


{{{16x^2-64=0}}} Start with the given equation.



Notice that the quadratic {{{16x^2-64}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=16}}}, {{{B=0}}}, and {{{C=-64}}}



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



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



{{{x = ((0) +- sqrt( (0)^2-4(16)(-64) ))/(2(16))}}} Plug in  {{{A=16}}}, {{{B=0}}}, and {{{C=-64}}}



{{{x = (0 +- sqrt( 0-4(16)(-64) ))/(2(16))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (0 +- sqrt( 0--4096 ))/(2(16))}}} Multiply {{{4(16)(-64)}}} to get {{{-4096}}}



{{{x = (0 +- sqrt( 0+4096 ))/(2(16))}}} Rewrite {{{sqrt(0--4096)}}} as {{{sqrt(0+4096)}}}



{{{x = (0 +- sqrt( 4096 ))/(2(16))}}} Add {{{0}}} to {{{4096}}} to get {{{4096}}}



{{{x = (0 +- sqrt( 4096 ))/(32)}}} Multiply {{{2}}} and {{{16}}} to get {{{32}}}. 



{{{x = (0 +- 64)/(32)}}} Take the square root of {{{4096}}} to get {{{64}}}. 



{{{x = (0 + 64)/(32)}}} or {{{x = (-0 - 64)/(32)}}} Break up the expression. 



{{{x = (64)/(32)}}} or {{{x =  (-64)/(32)}}} Combine like terms. 



{{{x = 2}}} or {{{x = -2}}} Simplify. 



So the solutions are {{{x = 2}}} or {{{x = -2}}} 
  


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