Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
If y=root(3y+4), then the product of all possible solution
[#permalink]
Show Tags
20 May 2010, 10:02
11
16
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 \(y\) cannot be negative as it equals to square root of some expression (\(\sqrt{expression}\geq{0}\)), so only one solution is valid \(y=4\).
If y=root(3y+4), then the product of all possible solution
[#permalink]
Show Tags
20 May 2010, 10:36
10
4
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 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 \(\geq{0}\)).
Re: If y=root(3y+4), then the product of all possible solution
[#permalink]
Show Tags
19 Apr 2013, 02:02
1
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 \(\frac{c}{a}\)
_________________
Did you find this post helpful?... Please let me know through the Kudos button.
Re: If y=root(3y+4), then the product of all possible solution
[#permalink]
Show Tags
06 Jul 2016, 23:44
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: \(y^2=3y+4\) --> \((y+1)(y-4)=0\) --> \(y=-1\) 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.
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.
_________________
Re: If y=root(3y+4), then the product of all possible solution
[#permalink]
Show Tags
07 Jul 2016, 00:09
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: \(y^2=3y+4\) --> \((y+1)(y-4)=0\) --> \(y=-1\) 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.
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.
Re: If y=root(3y+4), then the product of all possible solution
[#permalink]
Show Tags
07 Jul 2016, 00:27
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.
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. _________________
Re: If y=root(3y+4), then the product of all possible solution
[#permalink]
Show Tags
07 Jul 2016, 00:33
rahulforsure wrote:
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.
This applies to math generally. The square root cannot give negative result.
_________________
If y=root(3y+4), then the product of all possible solution
[#permalink]
Show Tags
Updated on: 25 Jul 2016, 22:31
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 (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
_________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.
Originally posted by LogicGuru1 on 07 Jul 2016, 01:09.
Last edited by LogicGuru1 on 25 Jul 2016, 22:31, edited 1 time in total.
Re: If y=root(3y+4), then the product of all possible solution
[#permalink]
Show Tags
16 Jul 2017, 13:54
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 _________________
"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."
Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475
close
44% (01:17) correct 
