Question 1086139
Use a substitution for ease,
{{{C=a-1/a}}}
Complete the square,
{{{x^2+Cx-1=0}}}
{{{(x^2+Cx+(C/2)^2)=1+(C/2)^2}}}
{{{(x+C/2)^2=1+C^2/4}}}
{{{x+C/2=0 +- sqrt((4+C^2)/4)}}}
{{{x=-C/2 +- sqrt(4+C^2)/2}}}
So then,
{{{C/2=a/2-1/(2a)}}}
{{{C/2=a^2/(2a)-1/(2a)}}}
{{{C/2=(a^2-1)/(2a)}}}
and
{{{C=(a^2-1)/a}}}
{{{C^2=(a^2-1)^2/a^2}}}
{{{4+C^2=4a^2/a^2+(a^2-1)^2/a^2}}}
{{{4+C^2=(4a^2+(a^2-1)^2)/a^2}}}
{{{4+C^2=(4a^2+a^4-2a^2+1)/a^2}}}
{{{4+C^2=(a^4+2a^2+1)/a^2}}}
{{{4+C^2=(a^2+1)^2/a^2}}}
So then substituting,
{{{2x=-C +- sqrt(4+C^2)}}}
{{{2x=(1-a^2)/a +- sqrt((a^2+1)^2/a^2)}}}
{{{2x=(1-a^2)/a +- (a^2+1)/a}}}
{{{2xa=1-a^2 +- (a^2+1)}}}
"Positive" solution:
{{{2xa=1-a^2 + a^2+1}}}
{{{2xa=2}}}
{{{highlight(x=1/a)}}}
"Negative" solution:
{{{2xa=1-a^2 - a^2-1)}}}
{{{2xa=-2a^2}}}
{{{highlight(x=-a)}}}