Difference between revisions of "2002 AMC 12A Problems/Problem 1"
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| − | {{duplicate|[[2002 AMC 12A Problems| | + | {{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #1]] and [[2002 AMC 10A Problems|2002 AMC 10A #10]]}} |
== Problem == | == Problem == | ||
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<math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math> | <math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math> | ||
| − | <math> \ | + | <math> \textbf{(A) } \frac{7}{2}\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 7\qquad \textbf{(E) } 13 </math> |
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| − | ===Solution 2== | + | ==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>\frac{14}4 = \boxed{\textbf{(A) }7/2}</math>. | |
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| + | ==Solution 2== | ||
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| + | We see a <math>(2x+3)</math> term in both addends so we can factor the equation to <math>(2x+3)((x-4)+(x-6))=(2x+3)(2x-10)=0.</math> The two roots become obvious: they are <math>-\frac{3}{2}</math> and <math>\frac{10}{2}.</math> Adding these two gives <math>\boxed{\textbf{(A) }7/2}.</math> | ||
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| + | -starwarmonkey | ||
| + | |||
| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/kNEqKCJsOR4 | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=A|before=First Question|num-a=2}} | {{AMC12 box|year=2002|ab=A|before=First Question|num-a=2}} | ||
{{AMC10 box|year=2002|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2002|ab=A|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 15:22, 26 August 2024
- 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
Solution 1
We expand to get 
which is 
after combining like terms. Using the quadratic part of Vieta's Formulas, we find the sum of the roots is 
.
Solution 2
We see a 
term in both addends so we can factor the equation to 
The two roots become obvious: they are 
and 
Adding these two gives 
-starwarmonkey
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
See also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| 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 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
