If |x^2 – 6| = x, which of the following could be the value of x? (A)

Question

If  |x^2 – 6| = x , which of the following could be the value of  x ?
(A) –2
(B) 0
(C) 1
(D) 3
(E) 5

Source: Manhattan Advanced quant

Abhishek009 · Answer

Plug in the options and check -

|x^2 – 6| = x

Or,  |3^2 – 6| = 3

Or,  |9 – 6| = 3  , Hence Answer must be (D)

NHiromoto · Answer

This equation means x ≥ 0.
So,

x^2 - 6 = x
x^2 - x - 6 =0
(x - 3)(x + 2) = 0
Because x≥0,
x = 3

philipssonicare · Answer

chetan2u 
I tried using the critical value method but I'm getting invalid results.

|x^2-6|-x=0 
Critical value is 6

Modulus is -|| when <6
Modulus is +|| when 6≤

-(x^2-6)-x=0; -x^2+6-x=0; x^2+x-6=0; (x+3)(x-2)=0; x=-3 or 2 
Neither -3 or 2 are <6. Invalid!

(x^2-6)-x=0; x^2-6-x=0; (x-3)(x+2)=0; x=-2 or 3 
Neither -2 nor 3 are 6≤. Invalid!

Please help!

chetan2u · Answer

You are OK with your solution.

What did we start with.... x^2<6 
 Now x can be -3 or 2. 
 x=-3, x^2=9  is NOT less than 6....Invalid
 x=2, x^2=4<6 ......Valid

What did we start with.... x^2\geq 6 
 Now x can be -2 or 3. 
 x=-2, x^2=4  is NOT greater than 6....Invalid
 x=3, x^2=9\geq{6} ......Valid

So, two valid values : 2 and 3.

vani2795 · Answer

DishaAgarwal12 · Answer

|x2-6|= sq root ((x2-6)^2)
sq root ((x2-6)^2)=x

sq both sides 
(x2-6)^2-x^2=0
x^4+36-13x^2=0
hidden quadratic
x^2=a
solve

a^2-13a+36=0

a=9 or a=4
hence, x=+3,-3 or+2,-2

it can't be negative because of the |x2-6|=x
hence x>=0

therefore 3 or 2, and 2 is not there in option

Prep Toolkit