What is the value of y?

Question

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

(2)  y^2–2y–8 = 0

GGMU · Answer

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.

ashwini86 · Answer

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 ??

ShreyasCM · Answer

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.

Sent from my A114 using  GMAT Club Forum mobile app

GGMU · Answer

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

chesstitans · Answer

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

GGMU · Answer

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.

sahilvijay · Answer

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

Bunuel · Answer

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.

sahilvijay · Answer

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.

kunalsinghNS · Answer

can you explain why the RHS will always be positive ?

sahilvijay · Answer

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

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masoodtaxed · Answer

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

jindal2910 · Answer

square of anything(X) that comes out will be mode of X. so, needs to positive. can't be negative

exc4libur · Answer

(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.

Answer (A)

kajaldaryani46 · Answer

——-
So we need to always check whether the solutions are valid when there’s an equation given like in the Q here - specially with a radical. Is my understanding correct?

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Thelegend2631 · Answer

Sounds Vague to me .
X can be -11 and 11 as well, how come we ignore possibility of -11 and -7

Hope Gmat Does't call ambiguity as a trick . !

Bunuel   IanStewart   MathRevolution

V2richandu · Answer

How in statement #1,RHS is considered positive ? .√x^2 is |x| which means -x if x<0 and +x if x>0.
So how come √49 is +7 ? √49 can be written as √(+7)^2 or √(-7)^2 . Isn't it ?

So shouldnt the answer be E .?

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bumpbot · Answer

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What is the value of y?

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