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If y=root(3y+4), then the product of all possible solution [#permalink]
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20 May 2010, 10:48
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If \(y=\sqrt{3y+4}\), then the product of all possible solution(s) for y is: A. 4 B. 2 C. 0 D. 4 E. 6
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If y=root(3y+4), then the product of all possible solution [#permalink]
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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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20 May 2010, 11:12
But don't we consider negative sq. root as well?



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If y=root(3y+4), then the product of all possible solution [#permalink]
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Prax wrote: 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 nonnegative 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 nonnegative 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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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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20 May 2010, 11:39
Thanks a lot Bunuel. its absolutely clear



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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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rakeshd347 wrote: noboru wrote: If y=sqrt(3y+4) then the product of all possible solution(s) for y is
a 4 b 2 c 0 d 4 e 6 D is the correct answer. The two solutions are 4 and 1 then 1 doesn't satisfy the equation so the only solution is 4. D is correct. 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 \(\frac{c}{a}\)
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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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12 May 2014, 02:13
Prax wrote: Can you please help me with this question:
If \(y=\sqrt{3y+4}\), then the product of all possible solution(s) for y is:
A. 4 B. 2 C. 0 D. 4 E. 6 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(y4) + 1(y  4) = 0 (y4) (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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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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07 Jul 2016, 00:44
Bunuel wrote: Prax wrote: Can you please help me with this question:
If \(y=\sqrt{3y+4}\), then the product of all possible solution(s) for y is: 4 2 0 4 6 Square both sides: \(y^2=3y+4\) > \((y+1)(y4)=0\) > \(y=1\) 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. 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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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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07 Jul 2016, 01:09
rahulforsure wrote: Bunuel wrote: Prax wrote: Can you please help me with this question:
If \(y=\sqrt{3y+4}\), then the product of all possible solution(s) for y is: 4 2 0 4 6 Square both sides: \(y^2=3y+4\) > \((y+1)(y4)=0\) > \(y=1\) 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. 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. Please read the whole thread: ifyroot3y4thentheproductofallpossiblesolution94527.html#p727486\(\sqrt{1}=1\), not 1 and 1.
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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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07 Jul 2016, 01:27
Bunuel wrote: rahulforsure wrote: Bunuel wrote: 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. Please read the whole thread: ifyroot3y4thentheproductofallpossiblesolution94527.html#p727486\(\sqrt{1}=1\), not 1 and 1. 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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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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If y=root(3y+4), then the product of all possible solution [#permalink]
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07 Jul 2016, 02:09
Prax wrote: Can you please help me with this question: If \(y=\sqrt{3y+4}\), then the product of all possible solution(s) for y is:
A. 4 B. 2 C. 0 D. 4 E. 6 It's a quadratic equation (y4)(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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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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07 Jul 2016, 05:22
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



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If y=root(3y+4), then the product of all possible solution [#permalink]
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15 Jan 2017, 07:02
Thanks for the feedback. I too am having a hard time understanding why the ultimate square root of 1 is not acceptable regardless if a negative was in the equation to get there. I understand Y can't be a number greater than 4/3 because that would result in a negative under the square root sign, but if the result is positive what is the harm? I think of the quadratic equation with the square root of b^24(a)(c) and several times the A or C may have a negative sign but it can still result in a positive answer. Can you please clear up this confusion?



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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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15 Jan 2017, 08:55
spencer4hire wrote: Thanks for the feedback. I too am having a hard time understanding why the ultimate square root of 1 is not acceptable regardless if a negative was in the equation to get there. I understand Y can't be a number greater than 4/3 because that would result in a negative under the square root sign, but if the result is positive what is the harm? I think of the quadratic equation with the square root of b^24(a)(c) and several times the A or C may have a negative sign but it can still result in a positive answer. Can you please clear up this confusion? This is addressed in the posts above. Please read the discussion above.
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Re: If y=root(3y+4), then the product of all possible solution [#permalink]
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