AMC 10 · 2006 · #7
Grade 8 arithmeticWhich of the following is equivalent to 1−xx−1x when x<0?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Simplify the expression $\sqrt{\dfrac{x}{1-\frac{x-1}{x}}}$ into one of the listed forms, given that $x$ is negative.
Givens: The expression $\sqrt{\dfrac{x}{1-\frac{x-1}{x}}}$; $x < 0$
Unknowns: Which answer choice the expression is equal to for every negative $x$
Understand
Restated: Simplify the expression $\sqrt{\dfrac{x}{1-\frac{x-1}{x}}}$ into one of the listed forms, given that $x$ is negative.
Givens: The expression $\sqrt{\dfrac{x}{1-\frac{x-1}{x}}}$; $x < 0$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #6 Guess and Check, #3 Eliminate Possibilities
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.
Execute — Answer: A
7.NS.A.1 Step 1 Clean up the bottom
- Look only at the denominator $1-\dfrac{x-1}{x}$.
- Write $1$ as $\dfrac{x}{x}$ so both parts share the denominator $x$, then subtract the numerators: $x-(x-1)=x-x+1=1$.
- So the whole bottom collapses to $\dfrac{1}{x}$.
💡 Give the two pieces the same denominator and the top almost cancels, leaving just $\tfrac{1}{x}$.
6.NS.A.1 Step 2 Do the division
- Now the inside of the root is $\dfrac{x}{1/x}$.
- Dividing by a fraction means multiplying by its reciprocal, so dividing by $\tfrac{1}{x}$ is the same as multiplying by $x$.
- That gives $x\cdot x=x^2$.
💡 Dividing by $\tfrac{1}{x}$ flips it to $x$, so the messy fraction becomes plain $x^2$.
8.EE.A.2 Step 3 Take the square root
- The expression is now $\sqrt{x^2}$.
- The square root of a squared number is its distance from zero, which is the absolute value: $\sqrt{x^2}=|x|$.
- This is always a number that is zero or positive, no matter the sign of $x$.
💡 A square erases the sign, so its root comes back as the size of $x$, never negative.
6.NS.C.7 Step 4 Use that x is negative
- For a negative number, the absolute value is the opposite of the number: $|x|=-x$, and since $x<0$ the quantity $-x$ is positive, matching the positive root.
- Testing $x=-2$ confirms it: the original expression gives $\sqrt{4}=2$, and $-x=-(-2)=2$.
- The look-alikes fail: $x=-2$, $1$, and $\sqrt{-2}$ do not equal $2$.
- So the answer is (A).
💡 The root must come out positive, and for a negative $x$ it is $-x$ that flips to positive.
7.NS.A.1 Look only at the denominator $1-\dfrac{x-1}{x}$. Write $1$ as $\dfrac{x}{x}$ so 6.NS.A.1 Now the inside of the root is $\dfrac{x}{1/x}$. Dividing by a fraction means mul 8.EE.A.2 The expression is now $\sqrt{x^2}$. The square root of a squared number is its d 6.NS.C.7 For a negative number, the absolute value is the opposite of the number: $|x|=-x Review
Reasonableness: 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) $x\sqrt{-1}$ is not even a real number, so neither can be the value of a real root.
Alternative: Skip most of the algebra and just test one safe value such as $x=-2$. The expression evaluates to $\sqrt{4}=2$, and among the choices only $-x=2$ matches; $x=-2$, $1$, and $\sqrt{-2}$ all miss. Using $x=-1$ would be a trap, because then both $-x$ and $1$ equal $1$.
CCSS standards used (min grade 8)
7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Combining $1-\frac{x-1}{x}$ over a common denominator down to $\frac{1}{x}$)6.NS.A.1Interpret and compute quotients of fractions and solve word problems (Dividing $x$ by $\frac{1}{x}$ by multiplying by the reciprocal to get $x^2$)8.EE.A.2Use square root and cube root symbols to represent solutions (Recognizing $\sqrt{x^2}=|x|$ as a nonnegative value)6.NS.C.7Understand ordering and absolute value of rational numbers (Turning $|x|$ into $-x$ because $x$ is negative)
⭐ The square root of $x^2$ is the size of $x$, so when $x$ is negative the answer is $-x$, which comes out positive.
⭐ The square root of $x^2$ is the size of $x$, so when $x$ is negative the answer is $-x$, which comes out positive.
More like this
Same archetype — closest grade level first.