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

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2003 AMC 10A Problem 5

Attempt Problem 5 of the 2003 AMC 10A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

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

$-\dfrac{5}{2}$

$0$

$3$

$5$

$6$

Answer: B

Concepts:quadratic factoring

Difficulty rating: 1200

Small Hint:

Expand $(d-1)(e-1)$ in terms of $d+e$ and $de$

Big Hint:

Use Vieta’s formulas to find $d+e$ and $de$ without solving for the roots

Solution:

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

Since one root equals $1,$ one of the two factors $d-1$ and $e-1$ equals $0,$ making the product $0.$

Thus, the correct answer is B.

Problem 5 in Other Years

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