Question 536619


{{{x^2-5=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=0}}}, and {{{C=-5}}}



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(1)(-5) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=0}}}, and {{{C=-5}}}



{{{x = (0 +- sqrt( 0-4(1)(-5) ))/(2(1))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (0 +- sqrt( 0--20 ))/(2(1))}}} Multiply {{{4(1)(-5)}}} to get {{{-20}}}



{{{x = (0 +- sqrt( 0+20 ))/(2(1))}}} Rewrite {{{sqrt(0--20)}}} as {{{sqrt(0+20)}}}



{{{x = (0 +- sqrt( 20 ))/(2(1))}}} Add {{{0}}} to {{{20}}} to get {{{20}}}



{{{x = (0 +- sqrt( 20 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (0 +- 2*sqrt(5))/(2)}}} 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 = (0)/(2) +- (2*sqrt(5))/(2)}}} Break up the fraction.  



{{{x = "" +- sqrt(5)}}} Reduce.  



{{{x = sqrt(5)}}} or {{{x = -sqrt(5)}}} Break up the expression.  



So the solutions are {{{x = sqrt(5)}}} or {{{x = -sqrt(5)}}}



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