AMC 10 · 2014 · #20
Grade 8 arithmeticFor how many integers x is the number x4−51x2+50 negative?
Pick an answer.
AMC 10 2014 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: Count the integers $x$ for which the value of $x^4-51x^2+50$ is a negative number.
Givens: The expression $x^4-51x^2+50$; Only the sign of the value matters — we need it to be less than $0$; $x$ must be an integer; Answer choices: (A) $8$, (B) $10$, (C) $12$, (D) $14$, (E) $16$
Unknowns: How many integer values of $x$ make the expression negative
Understand
Restated: Count the integers $x$ for which the value of $x^4-51x^2+50$ is a negative number.
Givens: The expression $x^4-51x^2+50$; Only the sign of the value matters — we need it to be less than $0$; $x$ must be an integer; Answer choices: (A) $8$, (B) $10$, (C) $12$, (D) $14$, (E) $16$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
The expression only uses $x^4$ and $x^2$, so letting $y=x^2$ (Tool #4) turns a scary quartic into an ordinary quadratic $y^2-51y+50$ that factors easily. Tool #7 (Identify Subproblems) then splits the job: first factor, then decide when the product is negative, then translate that back into a range for $x$. Tool #2 (Make a Systematic List) finishes by counting the integers in that range, remembering each positive value has a matching negative one.
Execute — Answer: C
6.EE.B.6 Step 1 Substitute to get a quadratic
- Every power of $x$ in the expression is even, so replace $x^2$ with a single letter $y$.
- Then $x^4=(x^2)^2=y^2$, and the expression becomes $y^2-51y+50$.
- The quartic in $x$ is now a familiar quadratic in $y$.
💡 Naming the repeated piece $x^2$ as one variable hides the scary exponent and leaves an ordinary quadratic.
6.EE.A.3 Step 2 Factor and undo the substitution
- Find two numbers that multiply to $50$ and add to $51$: those are $1$ and $50$.
- So $y^2-51y+50=(y-1)(y-50)$.
- Putting $y=x^2$ back in gives $(x^2-1)(x^2-50)$.
- A quick check by expanding confirms $x^4-50x^2-x^2+50=x^4-51x^2+50$.
💡 Factoring rewrites the sum-shaped expression as a product, and a product's sign is easy to control.
7.NS.A.2 Step 3 Decide when the product is negative
- A product of two numbers is negative only when one factor is positive and the other is negative.
- Compare the two factors: $x^2-1$ is always larger than $x^2-50$ (subtracting $1$ takes less away than subtracting $50$).
- So the larger one, $x^2-1$, must be the positive factor and the smaller one, $x^2-50$, the negative factor.
- That means $x^2-1>0$ and $x^2-50<0$ at the same time.
💡 For a product to dip below zero the factors must fight — one pushing positive, one pushing negative.
8.EE.A.2 Step 4 Turn the two conditions into a range
- The two conditions say $x^2>1$ and $x^2<50$, so $x^2$ must land strictly between $1$ and $50$.
- Since $7^2=49$ is still under $50$ but $8^2=64$ is over, the size of $x$ can be $2,3,4,5,6,$ or $7$ (a size of $1$ gives $x^2=1$, which is not greater than $1$).
💡 The perfect squares strictly between $1$ and $50$ are $4,9,16,25,36,49$, which pin down exactly which whole-number sizes of $x$ work.
6.EE.B.5 Step 5 Count both signs
- Each allowed size gives two integers, one positive and one negative, because the expression treats $x$ and $-x$ the same.
- The positive ones are $2,3,4,5,6,7$ and the negative ones are $-2,-3,-4,-5,-6,-7$.
- That is $6+6=12$ integers, which is choice (C).
💡 Because only even powers appear, every positive solution is mirrored by its negative twin, so count one side and double it.
6.EE.B.6 Every power of $x$ in the expression is even, so replace $x^2$ with a single let 6.EE.A.3 Find two numbers that multiply to $50$ and add to $51$: those are $1$ and $50$. 7.NS.A.2 A product of two numbers is negative only when one factor is positive and the ot 8.EE.A.2 The two conditions say $x^2>1$ and $x^2<50$, so $x^2$ must land strictly between 6.EE.B.5 Each allowed size gives two integers, one positive and one negative, because the Review
Reasonableness: Test a boundary and an interior value. At $x=1$: $1-51+50=0$, not negative, correctly excluded. At $x=2$: $16-204+50=-138$, negative, correctly included. At $x=7$: $2401-2499+50=-48$, negative, included. At $x=8$: $4096-3264+50=882$, positive, correctly excluded. So the usable sizes really are $2$ through $7$, giving $12$ integers, matching (C).
Alternative: Think of $y=x^4-51x^2+50$ as an upward-opening curve in $x^2$ with roots at $x^2=1$ and $x^2=50$; it is below zero exactly between those roots, i.e. for $1<x^2<50$. Reading off integer $x$ with $2\le|x|\le7$ again gives $12$ values, confirming (C) rather than the off-by-a-few traps (B) $10$ or (D) $14$.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Substituting $y=x^2$ to rewrite the quartic $x^4-51x^2+50$ as the quadratic $y^2-51y+50$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring $y^2-51y+50=(y-1)(y-50)$ and back-substituting to $(x^2-1)(x^2-50)$.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Using that a product is negative only when its two factors have opposite signs.)8.EE.A.2Use square root and cube root symbols to represent solutions (Converting $1<x^2<50$ into the size range $2\le|x|\le7$ via perfect squares.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Listing the integers that satisfy $2\le|x|\le7$ and counting positive and negative twins.)
⭐ Rename $x^2$ as one letter, factor into two pieces, and the answer is just the integers whose square lands strictly between $1$ and $50$ — counted for both signs.
⭐ Rename $x^2$ as one letter, factor into two pieces, and the answer is just the integers whose square lands strictly between $1$ and $50$ — counted for both signs.
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