AMC 10 · 2006 · #7

Grade 8 arithmetic
fraction-arithmeticabsolute-valuesigned-square-root identify-subproblems ↑ Prerequisites: fraction-arithmetic
📏 Medium solution 💡 2 insights
Problem
Given a number xx with x<0x \lt 0, simplify the expression x1x1x\sqrt{\frac{x}{1-\frac{x-1}{x}}} and decide which of the answer choices it is equal to.

Pick an answer.

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

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The expression is a fraction stacked inside a square root, so break it into small pieces: first clean up the little denominator, then do the division, then take the root. Because x is negative, the last piece needs care with the sign. A test value like x=-2 then checks the result and rules out the look-alike choices.

1STEP 1

Clean up the bottom

Write 11 as xx\frac{x}{x} and subtract the numerators, x(x1)=1x-(x-1)=1, so the whole bottom collapses to 1x\frac{1}{x}.

1-(x-1)/x=x/x-(x-1)/x=(x-(x-1))/x=1/x
2STEP 2

Do the division

Dividing by 1x\frac{1}{x} is the same as multiplying by its reciprocal xx, so under the root sits xxx \cdot x, namely x2x^2.

x/1/x=x · x=x²
3STEP 3

Take the square root

x2\sqrt{x^2} is the distance of xx from zero, the absolute value x|x|, which is never negative.

√(x²)=|x|
4STEP 4

Use that x is negative

Since xx is negative, x|x| is its opposite x-x, which is positive; x=2x=-2 checks out, giving 22 both ways — choice (A).

x < 0 → |x|=-x
Answer
-x
The result -x is positive when x < 0, which is exactly what a real square root should give. Plugging in x=-2 makes the original expression equal 2 and -x equal 2, so they agree. Choice (B) x would be negative, and choice (E) x√(-1) is not even a real number, so neither can be the value of a real root.
💡Key takeaway

The square root of x² is the size of x, so when x is negative the answer is -x, which comes out positive.

  • Clean up the bottom
  • Do the division
  • Take the square root
  • Use that x is negative