What is the product of all roots of the equation x + 1 = x- x - 1 ?

Question

What is the product of all roots of the equation  \sqrt{x} + 1 = x - \sqrt{x} - 1 ?

A. 1
B. 2
C. 3
D. 4
E. 5

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hunain505 · Answer

Ans: B -> 2

1.  Simplify 
\sqrt{x} = x - \sqrt{x}
2\sqrt{x} - x = 0

2.  Square both side to eliminate root 
2x -  x^2  = 0

This is a two-degree equation.

3. We can use the property of discriminant to identify possible solutions to equation.

Context: discriminant is  b^2  - 4ac 
If discriminant < 0 , no real roots of equation
If discriminant = 0, one root of equation
If discriminant >0 , two real roots of equation

In our case a= 1, b= -2:
 (-2)^2  - 4(1)(0) -> 4

A positive number. Means there're two real roots or answers to this equation

raianshul · Answer

on solving the equation, you will get the equation x^2 - 8x +4 =0, and product of the two roots = C/A in the equation Ax^2+Bx+C= 0, so the answer is 4/1 i.e. D

gmatophobia · Answer

x  + 1 = x -  x  - 1

x -  x  - 1 -  x  - 1 = 0

x - 2 x  - 2 = 0

Take   x   = y

y^2 - 2y - 2 = 0

Roots of y =   +2 \pm  4+8 /2

y_1 =  +2 + 2 3 /2   =  1+ 3

y_2 =  +2 - 2 3 /2   =  1- 3

y_1 =  x_1  = 1+ 3

y_2 =  x_2  = 1- 3

x_1 = (1+ 3 )^2 = (4+2 3 )

x_2 = (1- 3 )^2 = (4-2 3 )

x_1 * x_2 = (4+2 3 ) * (4-2 3 )

x_1 * x_2 = 16 - 12 = 4

Option D

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Bunuel     chetan2u

Request you to please review the approach and let me know if its correct.

Actually, I have a question and would like your thoughts on the same. Prior to solving the question as presented above, I had considered only one possible value of the root. In other words, I had disregarded the possibility of

y_2 =  x_2  = 1- 3

Reason for doing so  : LHS of the equation  ( x_2 )  should ideally result in a non - negative value, while the RHS of the equation  1- 3   is a negative value. I understanding that squaring removes the invalid possibility, but I had rejected that possibility merely by observing the equation. Was it too early to conclude ?

However, if I take just one root ( 1+ 3  ), none of the option choices match.

Can you please let me know if am I missing something in my understanding ?

chetan2u · Answer

gmatophobia 
I don’t find this to be a question worth spending time on.

Reasons

#We talk of roots of quadratic equation in GMAT, but the degree of the equation is 1 here as power of x is 1. 
#Even if we write it as  (\sqrt{x})^2-2\sqrt{x}-2=0 , the product of roots is -2 as roots of quadratic equation =c/a=-2/1=-2.
The roots will not be x but  \sqrt{x} . 
#Next, as found by you too, the value of  \sqrt{x}  cannot be negative.

Flawed question. Best is to leave it aside and move on to better questions.

What is the product of all roots of the equation x + 1 = x- x - 1 ?

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