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2002 AMC 10A Problem 10

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2002 AMC 10A Problem 10AMC 10 Step by Step
Name:                             Date: September 13, 2026
Quadratics · difficulty 2 of 5 · about 2 min · position P1-10

Compute the sum of all the roots of (2x+3)(x−4)+(2x+3)(x−6)=0(2x+3)(x-4)+(2x+3)(x-6)=0

  1. A)

    72\frac{7}{2}

  2. B)

    44

  3. C)

    55

  4. D)

    77

  5. E)

    1313

Answer

A

Key insight

Both terms share the factor (2x+3); factor it out instead of expanding, and the two roots appear immediately.

Solution

Both products contain (2x+3)(2x+3), so factor it out:
(2x+3)[(x−4)+(x−6)]=(2x+3)(2x−10)=0.(2x+3)\big[(x-4)+(x-6)\big] = (2x+3)(2x-10) = 0.
The roots are x=−32x = -\tfrac{3}{2} and x=5x = 5. Their sum is −32+5=72-\tfrac{3}{2} + 5 = \tfrac{7}{2}.

The answer is (A) 72\boxed{\textbf{(A)}\ \frac{7}{2}}.

Why this works

When an equation is a sum of products with a shared factor, factoring beats expanding: it keeps the roots visible and avoids arithmetic. If expansion is unavoidable, Vieta's formula r1+r2=−b/ar_1 + r_2 = -b/a gives the root sum without solving.

Alternative approach

Expand: 4x2−14x−30=04x^2 - 14x - 30 = 0, so by Vieta the sum of the roots is 144=72\frac{14}{4} = \frac{7}{2}.

The trap

Expanding everything and making an arithmetic slip, or applying Vieta to the un-simplified coefficients incorrectly.

Common mistakes

  • Expanding everything and making an arithmetic slip, or applying Vieta to the un-simplified coefficients incorrectly.
  • Dividing by (2x+3)(2x+3) and losing the root x=−32x = -\tfrac{3}{2}, which gives the wrong sum 55.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta