Question 183200
Since we're supposing that "x^2+a^2x+a^2 factors into (x+a)^2", this means that 


{{{x^2+a^2x+a^2=(x+a)^2}}}



{{{x^2+a^2x+a^2=(x+a)^2}}} Start with the given equation.



{{{x^2+a^2x+a^2=x^2+2ax+a^2}}} FOIL the right side



{{{cross(x^2-x^2)+a^2x+a^2=cross(x^2-x^2)+2ax+a^2}}} Subtract {{{x^2}}} from both sides. Note: the {{{x^2}}} terms cancel out.



{{{a^2x+cross(a^2-a^2)=2ax+cross(a^2-a^2)}}} Subtract {{{a^2}}} from both sides. Note: the {{{a^2}}} terms cancel out.



So we're left with: 


{{{a^2x=2ax}}}



{{{(a^2*cross(x))/cross(x)=(2a*cross(x))/cross(x)}}} Divide both sides by "x". Once again, the "x" terms cancel



{{{a^2=2a}}} Simplify



{{{a^2-2a=0}}} Subtract 2a from both sides.





{{{a(a-2)=0}}} Factor the left side 



Now set each factor equal to zero:


{{{a=0}}} or  {{{a-2=0}}} 



{{{a=0}}} or  {{{a=2}}}    Now solve for "a" in each case



Note: since we want a value of "a" "that is not zero", this means that we'll ignore the value {{{a=0}}}


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Answer:


So the solution is {{{a=2}}}