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

(Solution)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
 +
===Solution 1===
 
Using factoring:  
 
Using factoring:  
  
Line 15: Line 16:
 
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>.  
  
Therefore the answer is <math>(-\frac{5}{2}-1)(1-1)=(-\frac{7}{2})(0)=0 \Rightarrow B</math>
+
Therefore the answer is <math>\left(-\frac{5}{2}-1\right)(1-1)=\left(-\frac{7}{2}\right)(0)=\boxed{\mathrm{(B)}\ 0}</math>
  
OR we can use sum and product.
+
===Solution 2===
 +
We can use the sum and product of a quadratic:
  
<math>(d-1)(e-1)=de-(d+e)+1 \Rightarrow</math>
+
<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>
<math>product-sum+1 \Rightarrow</math>
 
<math>c/a-(-b/a)+1 \Rightarrow</math>
 
<math>(b+c)/a+1 \Rightarrow</math>
 
<math>0 \Rightarrow B</math>
 
  
 
== See Also ==
 
== See Also ==

Revision as of 17:06, 31 July 2011

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:

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

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS