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

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If y=root(3y+4), then the product of all possible solution [#permalink] New post 20 May 2010, 09:48
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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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Re: If y=root(3y+4), then the product of all possible solution [#permalink] New post 20 May 2010, 10:02
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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)(y-4)=0 --> y=-1 or y=4, but y can not 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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Re: If y=root(3y+4), then the product of all possible solution [#permalink] New post 20 May 2010, 10:12
But don't we consider negative sq. root as well?
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Re: If y=root(3y+4), then the product of all possible solution [#permalink] New post 20 May 2010, 10:36
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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 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 can not 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] New post 20 May 2010, 10:39
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] New post 19 Apr 2013, 02:02
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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] New post 12 May 2014, 00:10
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Re: If y=root(3y+4), then the product of all possible solution [#permalink] New post 12 May 2014, 01:13
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(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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Re: If y=root(3y+4), then the product of all possible solution   [#permalink] 12 May 2014, 01:13
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