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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. _________________

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 ??
_________________

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.

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
_________________

Kudos if you like my post

Last edited by anurag16 on 24 Jun 2016, 23:04, edited 1 time in total.

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

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.
_________________

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 _________________

Give Kudos for correct answer and/or if you like the solution.

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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