Competition · AMC preparation · step 4 of 4

AMC 10 · 2011B · #19

Grade 8 algebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

absolute-valuequadratic-equationssigned-square-root convert-to-algebra ↑ Prerequisites: quadratic-equations

📏 Medium solution 💡 3 insights

Problem

An equation sets two square roots equal:

\sqrt{5|x|+8} = \sqrt{x^2-16}

. A real number

x

counts as a root only when both radicands stay non-negative. Find the product of all the roots.

Pick an answer.

(A)

-64

(B)

-24

(C)

-9

(D)

(E)

576

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The messy part is |x| sitting next to x². Since x² = |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.

1STEP 1

Square both sides

Both sides are square roots, so squaring is safe: $5|x|+8 = x^2-16$ , which rearranges to $x^2-5|x|-24=0$ .

√(5|x| + 8) = √(x² - 16) → 5|x| + 8 = x² - 16 → x² - 5|x| - 24 = 0

2STEP 2

Substitute for the absolute value

Since $x^2 = |x|^2$ , set $u = |x| \ge 0$ ; the equation turns into the plain quadratic $u^2-5u-24=0$ .

u = |x| ≥ 0, x² = |x|² = u² → u² - 5u - 24 = 0

3STEP 3

Solve the quadratic

The two numbers with product $-24$ and sum $-5$ are $-8$ and $3$ , so $(u-8)(u+3)=0$ : $u=8$ or $u=-3$ .

u² - 5u - 24 = (u - 8)(u + 3) = 0 → u = 8 or u = -3

4STEP 4

Drop the impossible value

An absolute value is never negative, so $u=-3$ is impossible and only $|x|=8$ survives.

u = |x| ≥ 0 → u ≠ -3, so |x| = 8

5STEP 5

Check roots and multiply

So $x=8$ or $x=-8$ ; each checks out as $\sqrt{48}=\sqrt{48}$ , and their product is $-64$ , choice (A).

x = ± 8: √(48) = √(48) ✓ → 8 × (-8) = -64

Answer

-64

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 × 8 = 64 pins it to -64. Both roots also pass the original equation, giving √(48) = √(48).

💡Key takeaway

When an equation mixes |x| with x², rename |x| as one letter to get a plain quadratic — then throw out any answer that makes an absolute value negative.

Square both sides
Substitute for the absolute value
Solve the quadratic
Drop the impossible value
Check roots and multiply

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Square both sides

Substitute for the absolute value

Solve the quadratic

Drop the impossible value

Check roots and multiply

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