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2003 AMC 10A Problem 5

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2003 AMC 10A Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

Quadratics · difficulty 1 of 5 · about 1 min · position P1-10

Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$ . What is the value of $(d-1)(e-1)$ ?

A)
$-\frac{5}{2}$
B)
$0$
C)
$3$
D)
$5$
E)
$6$

Answer

Key insight

The coefficients 2 + 3 - 5 sum to zero, so x = 1 is a root and the factor (root - 1) kills the product.

Solution

Test $x = 1$ : $2(1)^2 + 3(1) - 5 = 0$ . So $1$ is a solution, meaning one of $d, e$ equals $1$ . Then one factor of $(d-1)(e-1)$ is $1 - 1 = 0$ , and the product is $0$ regardless of the other root.

(For completeness, $2x^2 + 3x - 5 = (x - 1)(2x + 5)$ , so the other root is $-\tfrac{5}{2}$ .)

The answer is $\boxed{\textbf{(B)}\ 0}$ .

Why this works

When a polynomial's coefficients sum to zero, $x = 1$ is a root. More generally, since $2x^2 + 3x - 5 = 2(x - d)(x - e)$ , substituting $x = 1$ gives $2(1-d)(1-e)$ , so $(d-1)(e-1)$ is just the polynomial's value at $1$ divided by the leading coefficient. Evaluating at a point is often faster than finding the roots.

Alternative approach

Vieta: $d + e = -\tfrac{3}{2}$ and $de = -\tfrac{5}{2}$ . Then $(d-1)(e-1) = de - (d+e) + 1 = -\tfrac{5}{2} + \tfrac{3}{2} + 1 = 0$ .

The trap

Using Vieta with the wrong signs, e.g. de + (d + e) + 1, and getting a nonzero value.

Common mistakes

Using Vieta with the wrong signs, e.g. de + (d + e) + 1, and getting a nonzero value.
Answering $-\tfrac{5}{2}$ (the product of the roots, choice (A)) instead of the requested expression.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Test small/specific values or special cases to find or verify the answer