AMC 10 · 2002 · #10
Grade 8 arithmeticCompute the sum of all the roots of
(2x+3)(x−4)+(2x+3)(x−6)=0
Pick an answer.
AMC 10 2002 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: Find the sum of every value of $x$ that makes $(2x+3)(x-4)+(2x+3)(x-6)=0$ true.
Givens: The equation $(2x+3)(x-4)+(2x+3)(x-6)=0$; It is written as two products added together, and both products share the factor $(2x+3)$; Answer choices: (A) $\frac{7}{2}$, (B) $4$, (C) $5$, (D) $7$, (E) $13$
Unknowns: The sum of all the roots (all the $x$ values that solve the equation)
Understand
Restated: Find the sum of every value of $x$ that makes $(2x+3)(x-4)+(2x+3)(x-6)=0$ true.
Givens: The equation $(2x+3)(x-4)+(2x+3)(x-6)=0$; It is written as two products added together, and both products share the factor $(2x+3)$; Answer choices: (A) $\frac{7}{2}$, (B) $4$, (C) $5$, (D) $7$, (E) $13$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Count the Complement
Instead of multiplying everything out, look at the shape of the left side: it is $(2x+3)$ times one thing plus $(2x+3)$ times another thing. Tool #16 (Change Focus) says read that structure rather than expanding, because the shared factor $(2x+3)$ can be pulled out. Once it is factored into a product that equals zero, tool #7 (Identify Subproblems) takes over: a product is zero exactly when one of its factors is zero, so the single hard equation splits into two easy linear equations. Solve each for one root, then add the roots.
Execute — Answer: A
6.EE.A.3 Step 1 Pull out the shared factor
- Both terms on the left have the same factor $(2x+3)$, so factor it out the same way $ab+ac=a(b+c)$ works.
- That leaves $(2x+3)\big[(x-4)+(x-6)\big]$.
- Combine the like terms inside the brackets: $(x-4)+(x-6)=2x-10$.
- The equation becomes $(2x+3)(2x-10)=0$.
💡 When every piece carries the same factor, you can grab that factor once instead of multiplying it back in.
8.EE.C.7 Step 2 Split into two linear equations
- A product of two things is zero only when at least one of them is zero.
- So $(2x+3)(2x-10)=0$ means $2x+3=0$ or $2x-10=0$.
- Solve each: from $2x+3=0$, $x=-\dfrac{3}{2}$; from $2x-10=0$, $x=5$.
- Those are the two roots.
💡 If two numbers multiply to zero, one of them had to be zero, so each factor hands you one root.
7.NS.A.1 Step 3 Add the two roots
- The question asks for the sum of the roots, so add them: $-\dfrac{3}{2}+5$.
- Write $5$ as $\dfrac{10}{2}$, then $-\dfrac{3}{2}+\dfrac{10}{2}=\dfrac{7}{2}$.
- The sum of all the roots is $\dfrac{7}{2}$, which is choice (A).
💡 Adding a negative fraction to a whole number is just lining them up over the same denominator.
6.EE.A.3 Both terms on the left have the same factor $(2x+3)$, so factor it out the same 8.EE.C.7 A product of two things is zero only when at least one of them is zero. So $(2x+ 7.NS.A.1 The question asks for the sum of the roots, so add them: $-\dfrac{3}{2}+5$. Writ Review
Reasonableness: The two roots are $-1.5$ and $5$, and $-1.5+5=3.5=\dfrac{7}{2}$, matching choice (A). A size check confirms the sign: one root is a small negative number and the other is $5$, so their sum should sit a bit below $5$ — exactly where $3.5$ lands. The equation is quadratic (its highest power is $x^2$), so two roots is the right count; there is no missing third root to add. Choice (D) $7$ is the trap for anyone who forgets to divide by the $2$ in each factor.
Alternative: Expand instead of factoring. $(2x+3)(x-4)=2x^2-5x-12$ and $(2x+3)(x-6)=2x^2-9x-18$; adding gives $4x^2-14x-30=0$. By Vieta's formulas the sum of the roots of $ax^2+bx+c=0$ is $-\dfrac{b}{a}=-\dfrac{-14}{4}=\dfrac{7}{2}$, the same answer (A) — and this way you never even solve for the individual roots.
CCSS standards used (min grade 8)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring the shared $(2x+3)$ out of both terms and combining $(x-4)+(x-6)$ into $2x-10$.)8.EE.C.7Solve linear equations in one variable (Solving $2x+3=0$ and $2x-10=0$ to get the roots $x=-\tfrac{3}{2}$ and $x=5$.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Adding the roots $-\tfrac{3}{2}+5=\tfrac{7}{2}$.)
⭐ When two terms share a factor, pull it out first — a product that equals zero breaks the problem into one easy equation per factor.
⭐ When two terms share a factor, pull it out first — a product that equals zero breaks the problem into one easy equation per factor.
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