Question 178395

{{{x^2+4x-5}}} Start with the given expression.



Take half of the {{{x}}} coefficient {{{4}}} to get {{{2}}}. In other words, {{{(1/2)(4)=2}}}.



Now square {{{2}}} to get {{{4}}}. In other words, {{{(2)^2=(2)(2)=4}}}



{{{x^2+4x+highlight(4-4)-5}}} Now add <font size=4><b>and</b></font> subtract {{{4}}}. Make sure to place this after the "x" term. Notice how {{{4-4=0}}}. So the expression is not changed.



{{{(x^2+4x+4)-4-5}}} Group the first three terms.



{{{(x+2)^2-4-5}}} Factor {{{x^2+4x+4}}} to get {{{(x+2)^2}}}.



{{{(x+2)^2-9}}} Combine like terms.



So after completing the square, {{{x^2+4x-5}}} transforms to {{{(x+2)^2-9}}}. So {{{x^2+4x-5=(x+2)^2-9}}}.



So {{{x^2+4x-5=0}}} is equivalent to {{{(x+2)^2-9=0}}}.



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Now let's solve {{{(x+2)^2-9=0}}}



{{{(x+2)^2-9=0}}} Start with the given equation.



{{{(x+2)^2=0+9}}}Add {{{9}}} to both sides.



{{{(x+2)^2=9}}} Combine like terms.



{{{x+2=0+-sqrt(9)}}} Take the square root of both sides.



{{{x+2=sqrt(9)}}} or {{{x+2=-sqrt(9)}}} Break up the "plus/minus" to form two equations.



{{{x+2=3}}} or {{{x+2=-3}}}  Take the square root of {{{9}}} to get {{{3}}}.



{{{x=-2+3}}} or {{{x=-2-3}}} Subtract {{{2}}} from both sides.



{{{x=1}}} or {{{x=-5}}} Combine like terms.



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Answer:



So the solutions are {{{x=1}}} or {{{x=-5}}}.