Question 57614
{{{sqrt(n-4)+ sqrt(n+4)=2sqrt(n-1)}}}
{{{(sqrt(n-4)+sqrt(n+4))^2=(2sqrt(n-1))^2}}}
{{{(sqrt(n-4)+sqrt(n+4))(sqrt(n-4)+sqrt(n+4))=2^2*(sqrt(n-1))^2}}}
{{{(sqrt(n-4))^2+sqrt((n+4)(n-4))+sqrt((n+4)(n-4))+(sqrt(n+4))^2=4(n-1)}}}
{{{(n-4)+2sqrt((n+4)(n-4))+(n+4)=4n-4}}}
{{{2n-4+4+2sqrt(n^2-16)=4n-4}}}
{{{2n+2sqrt(n^2-16)=4n-4}}}
{{{2n-2n+2sqrt(n^2-16)=4n-2n-4}}}
{{{2sqrt(n^2-16)=2n-4}}}
{{{2sqrt(n^2-16)/2=2n/2-4/2}}}
{{{sqrt(n^2-16)=n-2}}}
{{{(sqrt(n^2-16))^2=(n-2)^2}}}
{{{n^2-16=n^2-4n+4}}}
{{{n^2-n^2-16=n^2-n^2-4n+4}}}
{{{-16=-4n+4}}}
{{{-16-4=-4n+4-4}}}
{{{-20=-4n}}}
{{{-20/-4=-4n/-4}}}
{{{5=n}}}
:
Yopu must always check these kinds of equations for extraneous solutions by subtituting 5 in for n in the original equation.
{{{sqrt(5-4)+sqrt(5+4)=2*sqrt(5-1)}}}
{{{sqrt(1)+sqrt(9)=2*sqrt(4)}}}
{{{1+3=2*2}}}
{{{4=4}}} We've got a solution!!!
{{{highlight(n=5)}}}
Happy Calculating!!!