2002 AMC 12A Problems/Problem 1

The following problem is from both the 2002 AMC 12A #1 and 2002 AMC 10A #10, so both problems redirect to this page.

Problem

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

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

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 $\frac{14}4 = \boxed{\textbf{(A) }7/2}$.

Solution 2

We see a $(2x+3)$ term in both addends so we can factor the equation to $(2x+3)((x-4)+(x-6))=(2x+3)(2x-10)=0.$ The two roots become obvious: they are $-\frac{3}{2}$ and $\frac{10}{2}.$ Adding these two gives $\boxed{\textbf{(A) }7/2}.$

-starwarmonkey

Video Solution by Daily Dose of Math

https://youtu.be/kNEqKCJsOR4

~Thesmartgreekmathdude

See also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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 12 Problems and Solutions
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

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