Difference between revisions of "2002 AMC 10A Problems/Problem 10"

(fixed a mistake)
m (Solution 1)
Line 7: Line 7:
 
==Solution==
 
==Solution==
 
===Solution 1===
 
===Solution 1===
We expand to get <math>2x^2-8x+3x-12+2x^2-12x+3x-18=0</math> which is <math>4x^2-14x-30=0</math> after combining like terms. Using the quadratic part of [[Vieta's Formulas]], we find the sum of the roots is <math>\boxed{\text{(A)}\ 7/2}</math>.
+
We expand to get <math>2x^2-8x+3x-12+2x^2-12x+3x-18=0</math> which is <math>4x^2-14x-30=0</math> after combining like terms. Using the quadratic part of [[Vieta's Formulas]], we find the sum of the roots is <math>\frac{14}4 = \boxed{\text{(A)}\ 7/2}</math>.
  
 
===Solution 2===
 
===Solution 2===

Revision as of 12:52, 24 January 2009

Problem

What is the sum of all of the roots of $(2x + 3) (x - 4) + (2x + 3) (x - 6) = 0$?

$\text{(A)}\ 7/2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 13$

Solution

Solution 1

We expand to get $2x^2-8x+3x-12+2x^2-12x+3x-18=0$ which is $4x^2-14x-30=0$ after combining like terms. Using the quadratic part of Vieta's Formulas, we find the sum of the roots is $\frac{14}4 = \boxed{\text{(A)}\ 7/2}$.

Solution 2

Combine terms to get $(2x+3)\cdot\Big( (x-4)+(x-6) \Big) = (2x+3)(2x-10)=0$, hence the roots are $-\frac{3}{2}$ and $5$, thus our answer is $-\frac{3}{2}+5=\boxed{\text{(A)}\ 7/2}$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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