Printable

2006 AMC 10B Problem 7

Problem only Back to the problem

2006 AMC 10B Problem 7AMC 10 Step by Step

Name: Date: September 13, 2026

Exponents, Logarithms & Radicals · difficulty 2 of 5 · about 2 min · position P1-10

Which of the following is equivalent to $\sqrt{\frac{x}{1-\frac{x-1}{x}}}$ when $x < 0$ ?

A)
$-x$
B)
$x$
C)
$1$
D)
$\sqrt{\frac{x}{2}}$
E)
$x\sqrt{-1}$

Answer

Key insight

The denominator 1 - (x-1)/x collapses to 1/x, so the radicand is x^2 and the square root is |x|, which equals -x for negative x.

Solution

Simplify the inner denominator first:
$1 - \frac{x-1}{x} = \frac{x - (x-1)}{x} = \frac{1}{x}.$
Dividing $x$ by $\frac{1}{x}$ gives $x \cdot x = x^2$ , so the expression is $\sqrt{x^2}$ .

The principal square root of $x^2$ is $|x|$ , never just $x$ . Since the problem specifies $x < 0$ , we have $|x| = -x$ , which is a positive number.

The answer is $\boxed{\textbf{(A)}\ -x}$ .

Why this works

Clearing the nested fraction is routine; the real content is $\sqrt{x^2} = |x|$ . The hypothesis " $x < 0$ " is not decoration: it is exactly what decides between $x$ and $-x$ . Whenever a problem restricts the sign of a variable, expect an absolute value to be resolved by that sign.

Alternative approach

Test $x = -2$ : the denominator is $1 - \frac{-3}{-2} = 1 - \frac{3}{2} = -\frac{1}{2}$ , so the radicand is $\frac{-2}{-1/2} = 4$ and the root is $2 = -x$ . Avoid $x = -1$ , where (A) and (C) coincide.

The trap

Simplifying sqrt(x^2) to x and choosing (B); for x < 0 the principal square root is -x.

Common mistakes

Simplifying sqrt(x^2) to x and choosing (B); for x < 0 the principal square root is -x.
Testing only $x = -1$ , which cannot distinguish $-x$ from $1$ .

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer