AMC 10 · 2005 · #10
Grade 8 algebraThere are two values of a for which the equation 4x2+ax+8x+9=0 has only one solution for x. What is the sum of those values of a?
Pick an answer.
AMC 10 2005 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: The equation $4x^2+ax+8x+9=0$ has a parameter $a$. For exactly two special values of $a$, this equation has only one solution for $x$. Find the sum of those two values of $a$.
Givens: The equation is $4x^2+ax+8x+9=0$; There are exactly two values of $a$ that make it have only one solution for $x$; Answer choices: (A) $-16$, (B) $-8$, (C) $0$, (D) $8$, (E) $20$
Unknowns: The two values of $a$ that give only one solution; The sum of those two values of $a$
Understand
Restated: The equation $4x^2+ax+8x+9=0$ has a parameter $a$. For exactly two special values of $a$, this equation has only one solution for $x$. Find the sum of those two values of $a$.
Givens: The equation is $4x^2+ax+8x+9=0$; There are exactly two values of $a$ that make it have only one solution for $x$; Answer choices: (A) $-16$, (B) $-8$, (C) $0$, (D) $8$, (E) $20$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
First tidy the equation into standard form $4x^2+(a+8)x+9=0$. 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 $4x^2$ and $9$ are already the squares of $2x$ and $3$. Matching to the pattern $(2x\pm 3)^2=4x^2\pm 12x+9$ pins down the middle coefficient. Tool #4 (Introduce a Variable) then turns that match into the small equations $a+8=\pm 12$ and solves them for $a$; Tool #3 (Eliminate Possibilities) confirms the sum against the answer list.
Execute — Answer: A
7.EE.A.1 Step 1 Combine the two x-terms
- The two middle terms both contain $x$, so they are like terms and add together: $ax+8x=(a+8)x$.
- The equation becomes the standard quadratic form $4x^2+(a+8)x+9=0$.
- Only the middle coefficient depends on $a$; the ends $4$ and $9$ are fixed.
💡 Terms with the same variable part combine, so scattered $x$-terms collapse into one.
8.EE.A.2 Step 2 One solution means perfect square
- A quadratic equal to $0$ has just one solution only when its left side is a perfect square, so both roots coincide.
- Here the ends are already perfect squares: $4x^2=(2x)^2$ and $9=3^2$.
- So the whole left side must equal $(2x+3)^2$ or $(2x-3)^2$ — the sign of the $3$ is what we do not yet know.
💡 A repeated root packs the quadratic into a single square, and $4x^2$ and $9$ announce which square to try.
6.EE.A.3 Step 3 Match the middle term
- Expand each candidate square using the distributive property: $(2x+3)^2=4x^2+12x+9$ and $(2x-3)^2=4x^2-12x+9$.
- The $x^2$-terms and constant terms already agree, so the two expressions are equal only when the middle coefficients agree.
- That forces $a+8=12$ or $a+8=-12$.
💡 Two equal quadratics must match coefficient by coefficient, so the middle terms line up.
8.EE.C.7 Step 4 Solve for each value of a
- Each match is a one-step linear equation.
- From $a+8=12$, subtract $8$ to get $a=4$.
- From $a+8=-12$, subtract $8$ to get $a=-20$.
- These are exactly the two values that make the original equation have a single solution for $x$.
💡 Undo the $+8$ once for each sign to peel off $a$.
7.NS.A.1 Step 5 Add the two values
- The question asks for the sum of the two values of $a$.
- Add them: $4+(-20)=-16$.
- Scanning the answer list, $-16$ is choice (A), and no other choice matches, so (A) is the answer.
💡 The two values sit symmetrically around $-8$, so their sum lands at $2\times(-8)=-16$.
7.EE.A.1 The two middle terms both contain $x$, so they are like terms and add together: 8.EE.A.2 A quadratic equal to $0$ has just one solution only when its left side is a perf 6.EE.A.3 Expand each candidate square using the distributive property: $(2x+3)^2=4x^2+12x 8.EE.C.7 Each match is a one-step linear equation. From $a+8=12$, subtract $8$ to get $a= 7.NS.A.1 The question asks for the sum of the two values of $a$. Add them: $4+(-20)=-16$. Review
Reasonableness: The two values $a=4$ and $a=-20$ are the $\pm 12$ shifts of $-8$, so their sum must be $2\times(-8)=-16$ regardless of the exact shift — a good sign the algebra is consistent. Checking $a=4$: the equation becomes $4x^2+12x+9=(2x+3)^2=0$, whose only solution is $x=-\tfrac{3}{2}$. Checking $a=-20$: it becomes $4x^2-12x+9=(2x-3)^2=0$, whose only solution is $x=\tfrac{3}{2}$. Both genuinely have one solution, and $4+(-20)=-16$ matches choice (A).
Alternative: Use the discriminant. In $4x^2+(a+8)x+9=0$ one solution means the discriminant is $0$: $(a+8)^2-4\cdot 4\cdot 9=0$, i.e. $(a+8)^2=144$. Expanding gives $a^2+16a-80=0$, a quadratic whose two roots are the wanted values of $a$. By Vieta's formulas their sum is $-\tfrac{16}{1}=-16$, the same answer without solving for each root.
CCSS standards used (min grade 8)
7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Combining the like terms $ax+8x$ into $(a+8)x$ to write the equation in standard quadratic form.)8.EE.A.2Use square root and cube root symbols to represent solutions (Recognizing $4x^2=(2x)^2$ and $9=3^2$ so the one-solution left side must be the perfect square $(2x\pm 3)^2$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(2x\pm 3)^2$ to $4x^2\pm 12x+9$ and matching coefficients to get $a+8=\pm 12$.)8.EE.C.7Solve linear equations in one variable (Solving $a+8=12$ and $a+8=-12$ to find $a=4$ and $a=-20$.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Adding the two values $4+(-20)=-16$ to get the requested sum.)
⭐ A quadratic has just one solution when its left side is a perfect square, so match it to $(2x\pm 3)^2$ and read off the middle term.
⭐ A quadratic has just one solution when its left side is a perfect square, so match it to $(2x\pm 3)^2$ and read off the middle term.
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