If y=root(3y+4), then the product of all possible solution

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

Bunuel · Accepted Answer

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.

Bunuel · Answer

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 {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 {125} =5  and  \sqrt {-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} ).

Hope it's clear.

Prax · Answer

But don't we consider negative sq. root as well?

Prax · Answer

Thanks a lot Bunuel. its absolutely clear

MacFauz · Answer

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}

PerfectScores · Answer

y has to be a non negative value as y = square root of something hence we can eliminate A and B.

squaring both sides we get:
y^2 - 3y - 4 = 0
y^2 - 4y + y - 4 = 0
y(y-4) + 1(y - 4) = 0
(y-4) (y+1) = 0
y can be equal to 4 or -1.

Now -1 is not possible as -1 is not equal to square root of -1. Hence answer is 4. (D)

DigitsnLetters · Answer

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.

Bunuel · Answer

Please read the whole thread: if-y-root-3y-4-then-the-product-of-all-possible-solution-94527.html#p727486 \sqrt{1}=1 , not 1 and -1.

DigitsnLetters · Answer

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.

Bunuel · Answer

This applies to math generally. The square root cannot give negative result.

LogicGuru1 · Answer

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

nikhilbansal08 · Answer

y^2 = 3y+4.. solving the quadratic we get two values for Y : -1 or 4

But as Y is a root of something it cant take the negative value, hence only solution is 4

Hence answer is D

ydmuley · Answer

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 :good

gmatzpractice · Answer

y^2=3y+4
y^2-3y-4=0
(y+1)(y-4)=0
y=-1,4
y can't be negative because it's the result of a square root

Answer D

EgmatQuantExpert · Answer

Solution

Given  
 • y=√(3y+4)

To find  
 • The product of all the possible values of y.

Approach and Working out

Squaring on all the sides of y=√(3y+4), we get:
 • y^2 = 3y +4
• y^2 -3y -4 =0
• y^2 – 4y + y -4 =0
• y(y-4) +1(y-4) =0
• y = -1 and 4

Since y is equal to the square root of one number, it cannot be negative.
Hence, y =4
Thus, option D is the correct answer.

Correct Answer: Option D

Abhishek009 · Answer

y=\sqrt{3y+4}

Or,  y^2 = 3y + 4

Or,  y^2 - 3y - 4 = 0

Or,  y = 4 , Answer will be (D)

GmatPoint · Answer

﻿Given that  y\ =\ \sqrt{\ 3y+4} 
﻿Squaring both the sides we have :
 ﻿y^2=\ 3y+4 
 ﻿y^2-3y-4\ =\ 0 
 ﻿\left(y-4\right)\left(y+1\right)\ =0 
﻿y takes the values of 4 and -1.
﻿But since  \sqrt{\ x^2}=\ \left|x\right| 
﻿y = -1 does not satsify this case.
﻿Hence the only possible value if y that satsifies is 4.
﻿Hence the answer is 4.

eps018us · Answer

What about Vieta's formula? Why not c/a which would get to -4? Thanks

Bunuel · Answer

Vieta’s formula gives the product of all roots of the equation, including any extraneous ones created when squaring both sides.

But in this case, y = –1 doesn’t satisfy the original equation, so it must be discarded. Only y = 4 is valid, so the correct product is just 4, not –4.

If y=root(3y+4), then the product of all possible solution

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If y=root(3y+4), then the product of all possible solution

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