Square both sides: \(y^2=3y+4\) --> \((y+1)(y-4)=0\) --> \(y=-1\) or \(y=4\), but \(y\) cannot be negative as it equals to square root of some expression (\(\sqrt{expression}\geq{0}\)), so only one solution is valid \(y=4\).

Answer: D.
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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.
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Hello rakeshd347.,

If the two solutions are 4 and -1, that would mean that they both satisfy the equation and that is why they are the solutions. The product of all possible solutions is hence 4*-1 = -4.

If it helps, in an equation \(ax^2 + bx + c = 0\), the sum of the roots is \(\frac{-b}{a}\) and the product of the roots is \(\frac{c}{a}\)

I don't understand why y = -1 is not a valid solution. Let's evaluate the equation with y = -1.
LHS = y = -1.
RHS = sqrt(3y+4) = sqrt(3*(-1)+4) = sqrt(-3+4) = sqrt(1) = -1 = LHS. Where's the problem? It perfectly satisfies the equation.

Just saying that for y = -1 the sqrt is not valid is incorrect coz for y = -1 the expression under sqrt equals 1 which is +ve. Hence y = -1 is valid solution and answer for this question should be -1*4 = -4.

Hope I'm clear. Justify if I'm not.
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Ok. Got it. But does this apply to DS as well. E.g., if the deduction of any statement boils down to, say, x= sqrt(4), then can we say that that statement would be sufficient since it would yield just one value (i.e. the principal root)?

I'm a bit confused coz I remember coming across questions where both +ve & -ve values have been considered. Please help clarify my confusion. Thanks in advance.
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It's a quadratic equation (y-4)(y+1)=0
The two roots y=4 and -1
Since y is coming out of a square root it cannot be -ve
Therefore the root y=-1 is not valid leaving us with only one root y=4
The answer is D
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If \(y=\sqrt{3y+4}\), then the product of all possible solution(s) for y is:

\(y=\sqrt{3y+4}\)

Squaring on both sides.

\(y^2 = 3y + 4\)

\(y^2 - 3y - 4 = 0\)

\((y - 4) (y + 1) = 0\)

\(y = 4\) or

\(y = -1\)

As per the given equation for y we know that the value was in Square Root

As the value of square root cannot be negative, there can be only value for y here and that is y =4

Hence, Answer is D

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Solution

Given

• y=√(3y+4)

To find

• The product of all the possible values of y.

Approach and Working out

Squaring on all the sides of y=√(3y+4), we get:

• y^2 = 3y +4
• y^2 -3y -4 =0
• y^2 – 4y + y -4 =0
• y(y-4) +1(y-4) =0
• y = -1 and 4

Since y is equal to the square root of one number, it cannot be negative.
Hence, y =4
Thus, option D is the correct answer.

Correct Answer: Option D
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