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What is the value of y?  [#permalink]

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Question Stats: 28% (01:56) correct 72% (01:55) wrong based on 515 sessions

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What is the value of y?

(1) $$3y-1=\sqrt{(8y^2-4y+9)}$$

(2) $$y^2–2y–8 = 0$$

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Re: What is the value of y?  [#permalink]

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3
Yes the ans is A... This is a brilliant question from Magoosh... Below is the OE..

Using a tried and true DS strategy, start with the easier statement, Statement #2.

Statement #2:

(y – 4)(y + 2) = 0

y = +4 or y = –2

Since there are two values of y, this statement, alone and by itself, is not sufficient.

Statement #1:

This is an equation with a radical. The radical is already isolated, so square both sides.

(3y – 1)2 = 8y2 – 4y + 9

9y2 – 6y + 1 = 8y2 – 4y + 9

y2 – 2y – 8 = 0

Lo and behold! We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2. The naïve conclusion would be—this statement says exactly the same thing as the other. That's incorrect, though, because we don't know whether both of these values are valid solutions, or whether one or more is an extraneous root. We need to test this in the original equation.

Test y = +4 on equation 3y-1=√(8y^2-4y+9)
LHS=RHS

Test y = –2 on equation 3y-1=√(8y^2-4y+9)

The LHS and RHS are not equal, so this does not check! This value, y = –2, is an extraneous root.

(NB: it's often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)

Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.

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Re: What is the value of y?  [#permalink]

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anurag16 wrote:
Yes the ans is A... This is a brilliant question from Magoosh... Below is the OE..

Using a tried and true DS strategy, start with the easier statement, Statement #2.

Statement #2:

(y – 4)(y + 2) = 0

y = +4 or y = –2

Since there are two values of y, this statement, alone and by itself, is not sufficient.

Statement #1:

This is an equation with a radical. The radical is already isolated, so square both sides.

(3y – 1)2 = 8y2 – 4y + 9

9y2 – 6y + 1 = 8y2 – 4y + 9

y2 – 2y – 8 = 0

Lo and behold! We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2. The naïve conclusion would be—this statement says exactly the same thing as the other. That's incorrect, though, because we don't know whether both of these values are valid solutions, or whether one or more is an extraneous root. We need to test this in the original equation.

Test y = +4 on equation 3y-1=√(8y^2-4y+9)
LHS=RHS

Test y = –2 on equation 3y-1=√(8y^2-4y+9)

The LHS and RHS are not equal, so this does not check! This value, y = –2, is an extraneous root.

(NB: it's often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)

Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.

Hi ,

Given : y=-2

placing the values you get : -7 = sqrt(49) , this is possible

y = 4 you get 11 = sqrt(121)

So how can statement A be sufficient ??
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Re: What is the value of y?  [#permalink]

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anurag16 wrote:
Yes the ans is A... This is a brilliant question from Magoosh... Below is the OE..

Using a tried and true DS strategy, start with the easier statement, Statement #2.

Statement #2:

(y – 4)(y + 2) = 0

y = +4 or y = –2

Since there are two values of y, this statement, alone and by itself, is not sufficient.

Statement #1:

This is an equation with a radical. The radical is already isolated, so square both sides.

(3y – 1)2 = 8y2 – 4y + 9

9y2 – 6y + 1 = 8y2 – 4y + 9

y2 – 2y – 8 = 0

Lo and behold! We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2. The naïve conclusion would be—this statement says exactly the same thing as the other. That's incorrect, though, because we don't know whether both of these values are valid solutions, or whether one or more is an extraneous root. We need to test this in the original equation.

Test y = +4 on equation 3y-1=√(8y^2-4y+9)
LHS=RHS

Test y = –2 on equation 3y-1=√(8y^2-4y+9)

The LHS and RHS are not equal, so this does not check! This value, y = –2, is an extraneous root.

(NB: it's often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)

Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.

How can you say that for value y=4
11 = √121 since the value can be +11 or -11 for √121

Moreover there are other values for y which satisfy the equation like 7 = √ 49.
So how can value for y be uniquely determined by both the equations put together hence
E. Both statements are not sufficient.

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Re: What is the value of y?  [#permalink]

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ashwini86 wrote:
anurag16 wrote:
Yes the ans is A... This is a brilliant question from Magoosh... Below is the OE..

Using a tried and true DS strategy, start with the easier statement, Statement #2.

Statement #2:

(y – 4)(y + 2) = 0

y = +4 or y = –2

Since there are two values of y, this statement, alone and by itself, is not sufficient.

Statement #1:

This is an equation with a radical. The radical is already isolated, so square both sides.

(3y – 1)2 = 8y2 – 4y + 9

9y2 – 6y + 1 = 8y2 – 4y + 9

y2 – 2y – 8 = 0

Lo and behold! We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2. The naïve conclusion would be—this statement says exactly the same thing as the other. That's incorrect, though, because we don't know whether both of these values are valid solutions, or whether one or more is an extraneous root. We need to test this in the original equation.

Test y = +4 on equation 3y-1=√(8y^2-4y+9)
LHS=RHS

Test y = –2 on equation 3y-1=√(8y^2-4y+9)

The LHS and RHS are not equal, so this does not check! This value, y = –2, is an extraneous root.

(NB: it's often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)

Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.

Hi ,

Given : y=-2

placing the values you get : -7 = sqrt(49) , this is possible

y = 4 you get 11 = sqrt(121)

So how can statement A be sufficient ??

Hi ashwini86,
y=-2 will give -7 on LHS and 7 on RHS
As the LHS will not be equal to RHS so -2 will not be a valid root.
So it will have only one root i.e. only one value of y=4
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Originally posted by GGMU on 24 Jun 2016, 21:52.
Last edited by GGMU on 24 Jun 2016, 23:04, edited 1 time in total.
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Re: What is the value of y?  [#permalink]

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anurag16 wrote:
ashwini86 wrote:
anurag16 wrote:
Yes the ans is A... This is a brilliant question from Magoosh... Below is the OE..

Using a tried and true DS strategy, start with the easier statement, Statement #2.

Statement #2:

(y – 4)(y + 2) = 0

y = +4 or y = –2

Since there are two values of y, this statement, alone and by itself, is not sufficient.

Statement #1:

This is an equation with a radical. The radical is already isolated, so square both sides.

(3y – 1)2 = 8y2 – 4y + 9

9y2 – 6y + 1 = 8y2 – 4y + 9

y2 – 2y – 8 = 0

Lo and behold! We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2. The naïve conclusion would be—this statement says exactly the same thing as the other. That's incorrect, though, because we don't know whether both of these values are valid solutions, or whether one or more is an extraneous root. We need to test this in the original equation.

Test y = +4 on equation 3y-1=√(8y^2-4y+9)
LHS=RHS

Test y = –2 on equation 3y-1=√(8y^2-4y+9)

The LHS and RHS are not equal, so this does not check! This value, y = –2, is an extraneous root.

(NB: it's often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)

Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.

Hi ,

Given : y=-2

placing the values you get : -7 = sqrt(49) , this is possible

y = 4 you get 11 = sqrt(121)

So how can statement A be sufficient ??

Hi ashwini86,
y=-2 will give -7 on LHS and 7 on RHS
As the LHS will not be equal to RHS so -2 will not be a valid root.
So it will have only one root i.e. only one value of y=4

Hi anurag 16,
Is the sqrt of a perfect square treated as positive value by default?
If yes then A can be the solution
ashwini86 wrote:
anurag16 wrote:
Yes the ans is A... This is a brilliant question from Magoosh... Below is the OE..

Using a tried and true DS strategy, start with the easier statement, Statement #2.

Statement #2:

(y – 4)(y + 2) = 0

y = +4 or y = –2

Since there are two values of y, this statement, alone and by itself, is not sufficient.

Statement #1:

This is an equation with a radical. The radical is already isolated, so square both sides.

(3y – 1)2 = 8y2 – 4y + 9

9y2 – 6y + 1 = 8y2 – 4y + 9

y2 – 2y – 8 = 0

Lo and behold! We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2. The naïve conclusion would be—this statement says exactly the same thing as the other. That's incorrect, though, because we don't know whether both of these values are valid solutions, or whether one or more is an extraneous root. We need to test this in the original equation.

Test y = +4 on equation 3y-1=√(8y^2-4y+9)
LHS=RHS

Test y = –2 on equation 3y-1=√(8y^2-4y+9)

The LHS and RHS are not equal, so this does not check! This value, y = –2, is an extraneous root.

(NB: it's often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)

Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.

Hi ,

Given : y=-2

placing the values you get : -7 = sqrt(49) , this is possible

y = 4 you get 11 = sqrt(121)

So how can statement A be sufficient ??

Sent from my A114 using GMAT Club Forum mobile app
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Re: What is the value of y?  [#permalink]

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Hi ShreyasCM
Suppose we have x^2=49 we will have two roots +7 and -7 because both these values will satisfy the equation.
However, if we talk about only sqrt49 as an individual identity we will only get +7 as the solution.

I hope it's clear now why we A option is sufficient.
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Re: What is the value of y?  [#permalink]

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well, there is a trap of y >= 1/3, right?
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Re: What is the value of y?  [#permalink]

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1)
(3y-1)= √(8y^2-4y+9)
squaring both sides
9y^2+1-6y = 8y^2-4y+9
y^2-2y-8=0
(y-4)(y+2) =0
y=4 or y=-2
but y= -2 not possible as RHS √ (8y^2-4y+9) will always be positive, but y=-2 makes LHS negative
so single deifinite ans y=4. so A or D

2) y^2-2y-8=0
(y-4)(y+2) =0
y=4 or y=-2
Multiple Ans => D eliminated

A is the Ans
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Re: What is the value of y?  [#permalink]

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chesstitans wrote:
well, there is a trap of y >= 1/3, right?

$$3y-1=\sqrt{(8y^2-4y+9)}$$

Yes, since 3y-1 equals to the even root of some expression, then it cannot be negative, thus 3y - 1 > = 0 or y >= 1/3.
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Re: What is the value of y?  [#permalink]

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chesstitans wrote:
well, there is a trap of y >= 1/3, right?

chesstitans

You can see my solution, No need to assume y>=1/3
just whatever value of y you calculate should make LHS positive.

Hope you find this useful.
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Re: What is the value of y?  [#permalink]

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sahilvijay wrote:
1)
(3y-1)= √(8y^2-4y+9)
squaring both sides
9y^2+1-6y = 8y^2-4y+9
y^2-2y-8=0
(y-4)(y+2) =0
y=4 or y=-2
but y= -2 not possible as RHS √ (8y^2-4y+9) will always be positive, but y=-2 makes LHS negative
so single deifinite ans y=4. so A or D

2) y^2-2y-8=0
(y-4)(y+2) =0
y=4 or y=-2
Multiple Ans => D eliminated

A is the Ans

can you explain why the RHS will always be positive ?
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Re: What is the value of y?  [#permalink]

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Kunal - RHS HAS A square root- which is always positive

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What is the value of y?  [#permalink]

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why can't we plug in 4 and -2 in statement 2 to check which one is correct as we did in statement 1?

GGMU wrote:
Yes the ans is A... This is a brilliant question from Magoosh... Below is the OE..

Using a tried and true DS strategy, start with the easier statement, Statement #2.

Statement #2:

(y – 4)(y + 2) = 0

y = +4 or y = –2

Since there are two values of y, this statement, alone and by itself, is not sufficient.

Statement #1:

This is an equation with a radical. The radical is already isolated, so square both sides.

(3y – 1)2 = 8y2 – 4y + 9

9y2 – 6y + 1 = 8y2 – 4y + 9

y2 – 2y – 8 = 0

Lo and behold! We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2. The naïve conclusion would be—this statement says exactly the same thing as the other. That's incorrect, though, because we don't know whether both of these values are valid solutions, or whether one or more is an extraneous root. We need to test this in the original equation.

Test y = +4 on equation 3y-1=√(8y^2-4y+9)
LHS=RHS

Test y = –2 on equation 3y-1=√(8y^2-4y+9)

The LHS and RHS are not equal, so this does not check! This value, y = –2, is an extraneous root.

(NB: it's often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)

Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.
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Re: What is the value of y?  [#permalink]

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square of anything(X) that comes out will be mode of X. so, needs to positive. can't be negative
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Re: What is the value of y?  [#permalink]

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GGMU wrote:
What is the value of y?

(1) $$3y-1=\sqrt{(8y^2-4y+9)}$$
(2) $$y^2–2y–8 = 0$$

(1) $$3y-1=\sqrt{(8y^2-4y+9)}$$: sufic.
$$(3y-1)^2=8y^2-4y+9…y^2–2y–8=0…(y-4)(y+2)=0…y=(4,-2)$$
$$\sqrt{(8y^2-4y+9)}≥0…3y-1≥0…3y≥1…y≥1/3…y=4$$

(2) $$y^2–2y–8 = 0…(y-4)(y+2)=0…y=(4,-2)$$: insufic. Re: What is the value of y?   [#permalink] 11 Sep 2019, 06:01
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