An algebra problem by Russell Garay

Lets take $\sqrt { 72+\sqrt { 72+\sqrt { 72+\sqrt { 72+.......... } } } } = x$

Then, $x = \sqrt {72+x}$

So, $x^2 = 72 + x$

Then, $x^2-x-72 = 0$

Solving it, we get $x = 9$ or $x = -8$ .

In the options, since -8 is not given, 9 is the answer.

my procedure was derived while i was answering it mentally so pardon the less mathematical more logical approach haha i first assumed what would be the square of 72. being the closest perfect square is 81 and 64. one would assume that the value of the square of 72 is a value that is 9>x>8 or 8<x<9, since the equation state that we add the square of 72 to 72 before squaring it and since its infinite it is logical that the value that you will add to 72 before squaring it is a value near to 9 because of constantly adding terms. therefore 72 + 9 = 81, squaring 81, i got 9

Let's say x = sqrt(72 + sqrt(72 + sqrt (72 + sqrt (72... So, x^2 = 72 + sqrt(72 + sqrt(72 + sqrt(72... Thus, x^2 - 72 = sqrt(72 + sqrt(72 + sqrt(72.. But we know that the long ass expression on the RHS is actually x. So we then have x^2 - 72 = x. We then have a quadraric equation x^2 - x -72 = 0. Since 72 is equal to 8 9, -72 can be equal to 8 times -9 or vice versa. We have -x in the eq, not +x. So choose -9 and 8. We can now rewrite the quadratic equation as (x - 9) (x + 8) = 0. Thus x = 9 or -8. We know that x must be +9 because a positive square root only has a positive answer.

Let's say x = sqrt(72 + sqrt(72 + sqrt (72 + sqrt (72... So, x^2 = 72 + sqrt(72 + sqrt(72 + sqrt(72... Thus, x^2 - 72 = sqrt(72 + sqrt(72 + sqrt(72.. But we know that the long ass expression on the RHS is actually x. So we then have x^2 - 72 = x. We then have a quadraric equation x^2 - x -72 = 0. Since 72 is equal to 8 9, -72 can be equal to 8 times -9 or vice versa. We have -x in the eq, not +x. So choose -9 and 8. We can now rewrite the quadratic equation as (x - 9) (x + 8) = 0. Thus x = 9 or -8. We know that x must be +9 because a positive square root only has a positive answer.