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Modulus of a number cannot be negative. A simple idea that somehow escapes our mind every now and then.

## Counting - Linear Equations

What is the number of real solutions of the equation x2 - 7|x| - 18 = 0?
1. 2
2. 4
3. 3
4. 1

Choice A. 2

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## Detailed Solution

Let us split this into two cases. Case 1, when x is greater than 0 and Case 2, when x is lesser than 0.

Case 1

x > 0. Now, |x| = x
x2 – 7x – 18 = 0
(x – 9) (x + 2) = 0
x is either –2 or +9.

Case 2

x < 0. Now, |x| = –x
x2 + 7x – 18 = 0
(x + 9) (x – 2) = 0
x is either –9 or +2.

However, in accordance with the initial assumption that x < 0, x can only be –9 (cannot be +2).

Hence, this equation has two roots: –9 and +9.

Alternatively, we can treat this as a quadratic in |x|, the equation can be written as |x|2 – 7 |x| – 18 = 0.
Or, (|x| – 9) (|x| + 2) = 0
|x| = 9 or –2. |x| cannot be –2.
|x| = 9, x = 9 or –9.

Correct Answer: There are 2 solutions.

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