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

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<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>
  
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== Solution ==
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== Solutions ==
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===Solution 1===
 
Using factoring:  
 
Using factoring:  
  
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So <math>d</math> and <math>e</math> are <math>-\frac{5}{2}</math> and <math>1</math>.  
 
So <math>d</math> and <math>e</math> are <math>-\frac{5}{2}</math> and <math>1</math>.  
  
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Therefore the answer is <math>(-\frac{5}{2}-1)(1-1)=(-\frac{7}{2})(0)=0 \Rightarrow B</math>
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Therefore the answer is <math>\left(-\frac{5}{2}-1\right)(1-1)=\left(-\frac{7}{2}\right)(0)=\boxed{\mathrm{(B)}\ 0}</math>
  
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OR we can use sum and product.
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===Solution 2===
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We can use the sum and product of a quadratic (a.k.a Vieta):
  
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<math>(d-1)(e-1)=de-(d+e)+1 \Rightarrow</math>
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<math>(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}</math>
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<math>product-sum+1 \Rightarrow</math>
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<math>c/a-(-b/a)+1 \Rightarrow</math>
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<math>(b+c)/a+1 \Rightarrow</math>
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===Solution 3===
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<math>0 \Rightarrow B</math>
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By inspection, we quickly note that <math>x=1</math> is a solution to the equation, therefore the answer is
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<math>(d-1)(e-1)=(1-1)(e-1)=\boxed{\mathrm{(B)}\ 0}</math>
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===Solution 4===
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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 ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

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

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