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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But don't we consider negative sq. root as well?

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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Thanks a lot Bunuel. its absolutely clear

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 \(\frac{c}{a}\)

y has to be a non negative value as y = square root of something hence we can eliminate A and B.

squaring both sides we get:
y^2 - 3y - 4 = 0
y^2 - 4y + y - 4 = 0
y(y-4) + 1(y - 4) = 0
(y-4) (y+1) = 0
y can be equal to 4 or -1.

Now -1 is not possible as -1 is not equal to square root of -1. Hence answer is 4. (D)
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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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Please read the whole thread: if-y-root-3y-4-then-the-product-of-all-possible-solution-94527.html#p727486

\(\sqrt{1}=1\), not 1 and -1.
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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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This applies to math generally. The square root cannot give negative result.
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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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y^2 = 3y+4.. solving the quadratic we get two values for Y : -1 or 4

But as Y is a root of something it cant take the negative value, hence only solution is 4

Hence answer is D

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

Did you like the answer? 1 Kudos Please
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y^2=3y+4
y^2-3y-4=0
(y+1)(y-4)=0
y=-1,4
y can't be negative because it's the result of a square root

Answer D

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

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

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

Or, \(y = 4\), Answer will be (D)
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Given that \(y\ =\ \sqrt{\ 3y+4}\)
Squaring both the sides we have :
\(y^2=\ 3y+4\)
\(y^2-3y-4\ =\ 0\)
\(\left(y-4\right)\left(y+1\right)\ =0\)
y takes the values of 4 and -1.
But since \(\sqrt{\ x^2}=\ \left|x\right|\)
y = -1 does not satsify this case.
Hence the only possible value if y that satsifies is 4.
Hence the answer is 4.
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