Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution)
 
(3 intermediate revisions by 3 users not shown)
Line 4: Line 4:
 
<math> \mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math>
 
<math> \mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math>
  
−
== Solution ==
+
== Solutions ==
 
===Solution 1===
 
===Solution 1===
 
Using factoring:  
 
Using factoring:  
Line 30: Line 30:
  
 
===Solution 4===
 
===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>
+
The form <math>(d-1)(e-1)</math> resembles the factored form for the quadratic, namely <math>(x-d)(x-e)</math> based on the information given. 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>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:37, 19 August 2023

Problem

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

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

Solutions

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 $(x-d)(x-e)$ based on the information given. Note putting in 1 for $x$ in the that quadratic immediately yields the desired expression. Thus, $2(1)^2+3(1)-5 = \boxed{\mathrm{(B)}\ 0}$

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_5&oldid=197186"