Difference between revisions of "2003 AMC 10A Problems/Problem 5"

Revision as of 20:48, 25 April 2022

Problem

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

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

Solution

Solution 1

Using factoring:

$2x^{2}+3x-5=0$

$(2x+5)(x-1)=0$

$x = -\frac{5}{2}$ or $x=1$

So $d$ and $e$ are $-\frac{5}{2}$ and $1$ .

Therefore the answer is $\left(-\frac{5}{2}-1\right)(1-1)=\left(-\frac{7}{2}\right)(0)=\boxed{\mathrm{(B)}\ 0}$

Solution 2

We can use the sum and product of a quadratic (a.k.a Vieta):

$(d-1)(e-1)=de-(d+e)+1 \implies\text{product}-\text{sum}+1 \implies \dfrac{c}{a}-\left(-\dfrac{b}{a}\right)+1 \implies \dfrac{b+c}{a}+1= \dfrac{5}{-5}+1=\boxed{\mathrm{(B)}\ 0}$

Solution 3

By inspection, we quickly note that $x=1$ is a solution to the equation, therefore the answer is

$(d-1)(e-1)=(1-1)(e-1)=\boxed{\mathrm{(B)}\ 0}$

Solution 4

The form $(d-1)(e-1)$ resembles the factored form for the quadratic, namely $(x-d)(x-e)$ . Note putting in 1 for $x$ in the that quadratic immediately yields the desired expression. Thus, $2(1)^2+3(1)-5 = \boxed{\mathrm{(B)}\ 0}$

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 28: / Line 28: @@
 <math>(d-1)(e-1)=(1-1)(e-1)=\boxed{\mathrm{(B)}\ 0}</math>
+===Solution 4===
+The form <math>(d-1)(e-1)</math> resembles the factored form for the quadratic, namely <math>(x-d)(x-e)</math>. Note putting in 1 for <math>x</math> in the that quadratic immediately yields the desired expression. Thus, <math>2(1)^2+3(1)-5 = \boxed{\mathrm{(B)}\ 0}</math>
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