AMC10 2003 A

AMC10 2003 A · Q5

AMC10 2003 A · Q5. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

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

设$d$和$e$是方程$2x^2 + 3x -5 = 0$的根。求$(d -1)(e -1)$的值。

(A) $-\frac{5}{2}$ $-\frac{5}{2}$

(B) 0 0

(D) 5 5

(E) 6 6

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Since $0 = 2x^2 + 3x - 5 = (2x + 5)(x - 1)$ we have $d = -\frac{5}{2}$ and $e = 1$. So $(d - 1)(e - 1) = 0$.

（B）由于 $0 = 2x^2 + 3x - 5 = (2x + 5)(x - 1)$，我们有 $d = -\frac{5}{2}$ 且 $e = 1$。所以 $(d - 1)(e - 1) = 0$。

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