AMC 10 · 2011 · #19
Grade 8 algebraWhat is the product of all the roots of the equation 5∣x∣+8=x2−16.
Pick an answer.
AMC 10 2011 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: An equation sets two square roots equal: $\sqrt{5|x| + 8} = \sqrt{x^2 - 16}$. Find every value of $x$ that makes it true, then multiply all of those roots together.
Givens: $\sqrt{5|x| + 8} = \sqrt{x^2 - 16}$; The unknown $x$ can be positive or negative, so $|x|$ appears.
Unknowns: The product of all real solutions $x$.
Understand
Restated: An equation sets two square roots equal: $\sqrt{5|x| + 8} = \sqrt{x^2 - 16}$. Find every value of $x$ that makes it true, then multiply all of those roots together.
Givens: $\sqrt{5|x| + 8} = \sqrt{x^2 - 16}$; The unknown $x$ can be positive or negative, so $|x|$ appears.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #15 Organize Information in More Ways, #3 Eliminate Possibilities, #6 Guess and Check
The messy part is $|x|$ sitting next to $x^2$. Since $x^2 = |x|^2$, letting a single variable stand for $|x|$ turns the whole thing into a plain quadratic. From there we solve, throw out the impossible value, and check for fake roots.
Execute — Answer: A
8.EE.A.2 Step 1 Square both sides
- Both sides are square roots, so both are $\ge 0$.
- Squaring is safe to try and removes the radicals: $5|x| + 8 = x^2 - 16$.
- Move everything to one side to get $x^2 - 5|x| - 24 = 0$.
💡 Squaring undoes a square root, so it clears both radicals at once.
6.EE.A.2 Step 2 Substitute for the absolute value
- Notice $x^2 = |x|^2$.
- Let $u = |x|$, which must be $\ge 0$.
- The equation becomes a clean quadratic in $u$: $u^2 - 5u - 24 = 0$.
💡 Renaming $|x|$ as one letter hides the absolute value and reveals an ordinary quadratic.
8.EE.C.7 Step 3 Solve the quadratic
- Factor $u^2 - 5u - 24 = 0$ by finding two numbers that multiply to $-24$ and add to $-5$: those are $-8$ and $+3$.
- So $(u - 8)(u + 3) = 0$, giving $u = 8$ or $u = -3$.
💡 Factoring splits the quadratic into two simple products, each easy to set to zero.
6.NS.C.7 Step 4 Drop the impossible value
- Here $u = |x|$, and an absolute value can never be negative.
- So $u = -3$ is impossible and only $u = 8$ survives, meaning $|x| = 8$.
💡 Distance from zero is never negative, so a negative value for $|x|$ is ruled out immediately.
6.EE.B.5 Step 5 Check roots and multiply
- From $|x| = 8$ the candidates are $x = 8$ and $x = -8$.
- Squaring was not reversible, so test them: $\sqrt{5\cdot 8 + 8} = \sqrt{48}$ and $\sqrt{8^2 - 16} = \sqrt{48}$ — both sides match, and $x = -8$ gives the same values.
- Both are genuine roots, so the product is $8 \times (-8) = -64$, choice $\textbf{(A)}$.
💡 Plugging candidates back in confirms they truly solve the original equation, not just the squared one.
8.EE.A.2 Both sides are square roots, so both are $\ge 0$. Squaring is safe to try and re 6.EE.A.2 Notice $x^2 = |x|^2$. Let $u = |x|$, which must be $\ge 0$. The equation becomes 8.EE.C.7 Factor $u^2 - 5u - 24 = 0$ by finding two numbers that multiply to $-24$ and add 6.NS.C.7 Here $u = |x|$, and an absolute value can never be negative. So $u = -3$ is impo 6.EE.B.5 From $|x| = 8$ the candidates are $x = 8$ and $x = -8$. Squaring was not reversi Review
Reasonableness: The two roots $8$ and $-8$ are negatives of each other, so their product is negative — that instantly rules out the positive choices $24$ and $576$. The magnitude $8 \times 8 = 64$ pins it to $-64$. Both roots also pass the original equation, giving $\sqrt{48} = \sqrt{48}$.
Alternative: Set $y = x^2$. Squaring gives $5|x| + 8 = y - 16$, but $|x| = \sqrt{y}$, so $y - 5\sqrt{y} - 24 = 0$; factoring in $\sqrt{y}$ gives $(\sqrt{y} - 8)(\sqrt{y} + 3) = 0$, so $\sqrt{y} = 8$, $x^2 = 64$, and $x = \pm 8$ — same result.
CCSS standards used (min grade 8)
8.EE.A.2Use square root symbols and evaluate square roots of perfect squares (Squaring both sides to remove the two radicals.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting $u = |x|$ to turn the equation into a quadratic.)8.EE.C.7Solve equations in one variable (Factoring and solving $u^2 - 5u - 24 = 0$.)6.NS.C.7Understand ordering and absolute value of rational numbers (Rejecting $u = -3$ because $|x|$ cannot be negative.)6.EE.B.5Understand solving an equation as finding the values that make it true (Checking $x = \pm 8$ in the original equation to rule out extraneous roots.)
⭐ When an equation mixes $|x|$ with $x^2$, rename $|x|$ as one letter to get a plain quadratic — then throw out any answer that makes an absolute value negative.
⭐ When an equation mixes $|x|$ with $x^2$, rename $|x|$ as one letter to get a plain quadratic — then throw out any answer that makes an absolute value negative.
More like this
Same archetype — closest grade level first.