Dinesh,

I think this is PS problem.

Can you please post the answer options?

y=(3y + 4)^1/2

y^2=3y+4

y^2-3y-4=0

(y-4)(y+1)=0

y=4,y=-1

Product of all possible solutions=4*(-1)=-4

Selango,
I am sorry but your answer is incorrect. The OA is +4.

The explanation says: When you square the two sides you introduce an ambiguous root. It MAY OR MAY NOT satisfy the equation. Try putting it "-1" in the original equation and you will see that you get

-1 = (-3 + 4)^1/2
which gives you -1 =1

Hence the only valid root is 4, which is the OA.

Yes I thought about this root.

I have doubt here.

y=(3y + 4)^1/2

y=4,y=-1

y=-1

-1=sqrt(1)

sqrt(1)=1 or -1.

y=4

4=sqrt(16)

sqrt(16)=4 or -4.

See these above values.y=-1 satisfy the above equation.

Am I missing something?

dinesh19aug wrote:Selango,
I am sorry but your answer is incorrect. The OA is +4.

The explanation says: When you square the two sides you introduce an ambiguous root. It MAY OR MAY NOT satisfy the equation. Try putting it "-1" in the original equation and you will see that you get

-1 = (-3 + 4)^1/2
which gives you -1 =1

Hence the only valid root is 4, which is the OA.

If the only valid root is 4, then how can you take the product of all possible roots? You can't take the product of 1 number (that's non-sensical).

When you say "MCAT 1", do you mean MGMAT CAT 1? Or is this from another source? MGMAT doesn't usually create invalid questions.

dinesh19aug wrote:Selango,
I am sorry but your answer is incorrect. The OA is +4.

The explanation says: When you square the two sides you introduce an ambiguous root. It MAY OR MAY NOT satisfy the equation. Try putting it "-1" in the original equation and you will see that you get

-1 = (-3 + 4)^1/2
which gives you -1 =1

Hence the only valid root is 4, which is the OA.

Did you copy the question exactly? Or did you add the "^1/2" because you couldn't copy the square root sign properly?

If the right side were "sqrt(3y + 4)", then it would only include the positive root. However, if it's reproduced correctly, then it should include both the positive and negative root.

Stuart Kovinsky wrote:

dinesh19aug wrote:Selango,
I am sorry but your answer is incorrect. The OA is +4.

The explanation says: When you square the two sides you introduce an ambiguous root. It MAY OR MAY NOT satisfy the equation. Try putting it "-1" in the original equation and you will see that you get

-1 = (-3 + 4)^1/2
which gives you -1 =1

Hence the only valid root is 4, which is the OA.

If the only valid root is 4, then how can you take the product of all possible roots? You can't take the product of 1 number (that's non-sensical).

When you say "MCAT 1", do you mean MGMAT CAT 1? Or is this from another source? MGMAT doesn't usually create invalid questions.

Yes this MGMAT cat question and I copied the question exactly as it was in the CAT.

I am not sure about this.

sqrt(1)=1 or -1

If you say -1 not satisfy the equation.lets try with y=4

4=sqrt(16)

sqrt(16)=4 or -4.Why should v take only 4 as its root?-4 also is valid.

Can you please post the answer options?

What's the OA!!!!!!!!!!! PILIZZ GIVE IT !!

hi guys
to me the answer is 4
my reasoning
for sure the given equation has two roots after squaring
y1=4, and y2=-1

but now insert y2=-1 in both parts of the equation and check equality
(3*(-1)+4)^1/2=(-3+4)^1/2=1
so right part of the equation=1
but left part still y=-1
-1=1 no! y=-1 is not valid root here ( to me that is the trap)
so we are left with y=4

I am not still convinced...sqrt(16) can be 4 or -4

Selango: I have already posted the answer as 4.
Still here are the choices
A) -4
B) 1
C) -1
D) 4
E) 8

The question does not say that there HAS to be two roots. The question says what is product of possible root(s). There could be 1 or 2 or more than 2 roots.

Clock60 very well explained. There can be no detailed explanation than this.

Q: If y=(3y + 4)^1/2 then the product of all possible solution(s) for y is?

I got this question in MCAT 1 and got it wrong. I read the solution and was able to understand the reasoning. I DO NOT have any doubts on this question. Just wanted to share the question with everyone on the forums.


Solution: the 1/2 on the R.H.S (Right Hand side) of the = sign when transferred to the L.H.S makes it

y² = 3Y+4 => y²-3Y-4 =0 (Product is 4 and the Sum is -3 so -4 &1 are your best options)

y²-4Y+Y-4=0 => y(Y-4) +1(Y-4)=0 => (Y-4) =0 ; (Y+1)=0 => Y= 4 or Y=-1

Substitute the values in the equation y²-3Y-4=0 => First take 4 => 4² -3*4 -4 =0 ; 16-12-4 =>16-16 =0

Take the second value "-1" => -1² -3*(-1) -4 =0 +> 1 + 3 -4 => 4-4 =0.

From the list of Answer option it should be either 4 or -1

Feel free to comment if the approach is incorrect.