2003 AMC 10A Problem 5 — Solution | LIVE by Po-Shen Loh

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2003 AMC 10A Problem 5

Below is the professionally curated solution for Problem 5 of the 2003 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10A solutions, or check the answer key.

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Concepts:quadratic factoring

Difficulty rating: 1200

5.

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

$-\dfrac{5}{2}$

$0$

$3$

$5$

$6$

Solution:

Factoring gives $2x^2 + 3x - 5 = (2x + 5)(x - 1),$ so the roots are $-\dfrac{5}{2}$ and $1.$

Since one root equals $1,$ the factor $(e - 1) = 0,$ making the product $0.$

Thus, the correct answer is B.

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Problem 5 in Other Years

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