What is the range of all the roots of |x^2 - 2| = x ?

In short, we will be solving for 2 quadratics. The two are:

1) x^2-x-2=0
2) x^2+x-2=0

The roots for quadratic 1 are {-2,1} and the roots for quadratic 2 are {2,-1}.

Range is highest - lowest, i.e. 2-(-2) =4

answer is A... or so I hope

Yeah you're right. Totally forgot to go back to the stem. -1 and -2 don't satisfy.

Attached image.
there are only 2 roots.
We can solve it algebrically as well.

Answer: C

I guess all the four integers, -2,-1,1 & 2 satisfies the equation mod (x^2-2) = X. Please let me know if I am missing some point. Though I understood the explanation by graph method, I am not able to understand the way you have explained it.

Hi All,

You have to be VERY careful with this question. It's actually built more to test your attention-to-detail than your "math skills"....

We're asked to find the RANGE of the roots of the following equation: |x^2 - 2| = x ?

Before you jump in and start doing calculations, there are a couple of points to note:

1) Absolute value calculations can NEVER equal a negative number. Here, we have an EQUATION set equal to X. As such, X CANNOT be NEGATIVE.
2) The answer choices are small integers, so the roots of the above equation are likely also integers that are relatively close to one another.
3) Since the question mentions ROOTS, there should be at least 2 solutions.

A bit of basic "brute force" is all that's really needed to find the roots of the equation....

Could X = 0?
|0-2| is NOT 0, so X cannot be 0

Could X = 1?
|1-2| does = 1, so X = 1 is a root

Could X = 2?
|4-2| does = 2, so X = 2 is a root

Could X = 3?
|9-2| is NOT 3, so X cannot be 3

As X gets bigger, the absolute value calculation gets even larger (and farther 'away') from X.

Thus, the only roots are 1 and 2. The range is 2-1 = 1

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Bunuel VeritasKarishma

I go the roots as x = 2, -1 and x = -2, 1.

I substituted them in the original equation to check if they are satisfying. -1, -2 are not satisfying.

But I have a doubt. The modulus value given in the original equation is x^2- 2. Without substituting the values in the original equation, if I consider that x^2-2 >=0 for verification purpose (because it is a modulus), then I get
=> x^2-2 >= 0
=> x^2 >= 2

In that case x=2 and x= -2 should satisfy this right? And 1, -1 should not satisfy.

Since the absolute value on the left side cannot be equal to a negative value, the right side of the equation must be NONNEGATIVE.
Thus, only nonnegative values for x are viable here.

Case 1: x²-2 = x
x² - x - 2 = 0
(x-2)(x+1) = 0.
x=2 or x=-1.
Since x must be nonnegative, only x=2 is viable.

Case 2: x²-2 = -x
x² + x - 2 = 0
(x+2)(x-1) = 0.
x=-2 or x=1.
Since x must be nonnegative, only x=1 is viable.

The range of the two roots = greater root - smaller root = 2-1 = 1.

Thank you for your response.

But my question is, as they can't be non-negative, after finding the roots, I considered that x^2-2 has to be >=0 from which x^2 >= 2.

The roots that satisfy this condition are 2 and -2.

The roots 1 and -1 does not satisfy this. How to decide which one to consider?

|x^2 - 2| cannot be negative but x^2 - 2 (expression in the modulus) can be. For example, if x = 1, then x^2 - 2 = -1 but | x^2 - 2| = |-1| = 1.

roots of |x^2 - 2| = x
x^2-x-2=0
x=2,+1
and x^2+x-2=0
x=-2,-1
-2,-1,1
x cannot be -ve as value is in mod
so 1,2 ; range 1
option D

x^2 - 2 = x
x^2 - x - 2
x = 2 and x = -1

x^2 - 2 = - x
x^2 + x - 2
x = - 2 and x = 1

Since we can't take negative values, the only two valid answers are 1 and 2.

2 - 1 = 1. Answer is D.

To see why negative values don't work, you can plug the negative into the initial equation:

|-2^2 - 2| = -2
2 is not equal to -2

Hi, I don't understand why -1 is not correct.
I plugged the numbers and this is what I found:

2^2-2=2
4-2=2
2=2 (2 is acceptable)

-1^2-2=-1
1-2= -1
-1=-1 (why -1 is wrong? -1^2 equal to 1 so the two values are the same)

I don't know what I did wrong

Hi saahulu,

The original equation that we are given to work with is:

|X^2 - 2| = X

This equation includes an ABSOLUTE VALUE that you did not include in either of your calculations. What happens when you plug X = -1 into this equation?

GMAT assassins aren't born, they're made,
Rich

Thank you!!!!! much more clear

I would square both sides to get rid of the absolute values and then put the roots back to check which ones satisfy the original equation.

\(|x^2 - 2|^2 = x^2\)

\(x^4 -5x^2 + 4 = 0\)

\((x^2 - 4) (x^2 - 1) = 0\)

x = 2, -2, 1, -1

Only 1 and 2 satisfy so range = 2 - 1 = 1

Think about why it is critical in this case to put back the roots in original equation to check. Why do we get extraneous roots?

Highlighted is not correct.

x^2 - 2 need not be positive. When you take its absolute value, it BECOMES positive even if it were negative.

If x = -1,
x^2 - 2 = 1 - 2 = -1
Here, x^2 - 2 is negative and there is nothing which conflicts with that.
Now, |x^2 - 2| = |-1| = 1
Now, if you want to equate it to x, you cannot because x = -1 and that is your problem. That is the reason -1 is not a root.


If x = 1,
x^2 - 2 = 1 - 2 = -1
Here, x^2 - 2 is negative and there is nothing which conflicts with that.
Now, |x^2 - 2| = |-1| = 1This is the same as x so you can equate it to x. Hence 1 is a root.

What is the range of all the roots of |x^2 - 2| = x?

\(\left|x^2-2\right|\ =\ x\)
Considering the function inside the modulus to be positive and negative :
If the function is positive :
This is possible if \(\left|x\right|\ \ge\ \sqrt{\ 2}\\
\)

\(x^2-x-2\ =\ 0\)
The roots of the equation are 2, -1
2 satisfies the condition.
Considering the function inside the modulus to be negative.
\(\left|x\right|<\sqrt{\ 2}\)
\(x^2+x-2\ =\ 0\)
x = -2, x = 1.
1 satisfies the condition.
The range of the roots : ( 2 - 1 ) = 1.