If y=root(3y+4), then the product of all possible solution

This issue was discussed several times lately on the forum and let me assure you: square root function cannot give negative result.

Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

So when we see \(y=\sqrt{3y+4}\) we can deduce TWO things:
A. \(y\geq{0}\) - as square root function cannot give negative result;
B. \(3y+4\geq{0}\) - as GMAT is dealing only with real numbers and even roots of negative number is undefined (\(3y+4\) is under square root so it must be \(\geq{0}\)).

Hope it's clear.