x^2-4=sqrt(x+4)
Hint: there is a tricky way to solve it !
(In reply to
re: Two methods, tricky and brute force by Ady TZIDON)
Generalize into x^2-t = sqrt(x+t) and then some algebra:
(x^2-t)^2 = (sqrt(x+t))^2
x^4 - 2x^2t + t^2 = x+t
t^2 - (2x^2+1)t + (x^4-x) = 0
And solve for t via quadratic formula:
t = ( 2x^2+1 +/- sqrt[(2x^2+1)^2-4*(x^4-x)] )/2
t = ( 2x^2+1 +/- sqrt[4x^2-4x+1] )/2
t = ( 2x^2+1 +/- (2x+1) )/2
t = x^2+x+1 OR t = x^2-x
And now to restore the original t=4:
4=x^2+x+1 OR 4=x^2-x
x^2+x-3=0 OR x^2-x-4=0
Then solving both quadratics the possible x are (-1+sqrt[13])/2, (-1-sqrt[13])/2, (1+sqrt[17])/2, (1-sqrt[17]).
But we still need to double check to see if squaring the equation at the beginning added extraneous roots. Two roots are then discarded leaving x=(-1-sqrt[13])/2 and (1+sqrt[17])/2 as the solutions.
re(2): Two methods, tricky and brute force - Ady's Trick
