AMC 10 · 2005 · #9

Grade 8 algebra
quadratic-equationsperfect-squares pattern-recognitionconvert-to-algebra ↑ Prerequisites: quadratic-equationsperfect-squarespolynomial-factoring
📏 Medium solution 💡 2 insights
Problem
A quadratic has a parameter in its middle term. For exactly two values of that parameter the equation has only one solution. Find the sum of those two values.

Pick an answer.

(A)
-16
(B)
-8
(C)
0
(D)
8
(E)
20
How to solve
Strategy Look for a Pattern

First tidy the equation into standard form 4x²+(a+8)x+9=0. The key is Tool #5 (Look for a Pattern): a quadratic set equal to 0 has only one solution exactly when its left side is a perfect square, and the outer terms 4x² and 9 are already the squares of 2x and 3. Matching to the pattern (2x± 3)²=4x²± 12x+9 pins down the middle coefficient. Tool #4 (Introduce a Variable) then turns that match into the small equations a+8=± 12 and solves them for a; Tool #3 (Eliminate Possibilities) confirms the sum against the answer list.

1STEP 1

Combine the two x-terms

Combining like terms puts the parameter into one coefficient.

4x²+ax+8x+9=4x²+(a+8)x+9=0
2STEP 2

One solution means perfect square

A single solution means the side is a perfect square.

4x²+(a+8)x+9=(2x± 3)²
3STEP 3

Match the middle term

The ends are already squares, so only the middle needs matching, with two signs.

(2x± 3)²=4x²± 12x+9 → a+8=± 12
4STEP 4

Solve for each value of a

Each sign gives one value, 4 and -20.

a+8=12→ a=4; a+8=-12→ a=-20
5STEP 5

Add the two values

Adding them gives -16, choice (A).

4+(-20)=-16 → (A)
Answer
-16
The two values a=4 and a=-20 are the ± 12 shifts of -8, so their sum must be 2×(-8)=-16 regardless of the exact shift — a good sign the algebra is consistent. Checking a=4: the equation becomes 4x²+12x+9=(2x+3)²=0, whose only solution is x=-3/2. Checking a=-20: it becomes 4x²-12x+9=(2x-3)²=0, whose only solution is x=3/2. Both genuinely have one solution, and 4+(-20)=-16 matches choice (A).
💡Key takeaway

A quadratic has just one solution when its left side is a perfect square, so match it to (2x± 3)² and read off the middle term.

  • Combine the two x-terms
  • One solution means perfect square
  • Match the middle term
  • Solve for each value of a
  • Add the two values