Difference between revisions of "2020 AMC 10A Problems/Problem 5"
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== Solution 2== | == Solution 2== | ||
| − | We have the equations <math>x^2-12x+32=0</math> and <math>x^2-12x+36= | + | We have the equations <math>x^2-12x+32=0</math> and <math>x^2-12x+36=0</math>. |
Notice that the second is a perfect square with a double root at <math>x=6</math>, and the first has real roots. By Vieta's, the sum of the roots of the first equation is <math>12</math>. <math>12+6=\boxed{\text{(C) }18}</math>. | Notice that the second is a perfect square with a double root at <math>x=6</math>, and the first has real roots. By Vieta's, the sum of the roots of the first equation is <math>12</math>. <math>12+6=\boxed{\text{(C) }18}</math>. | ||
Revision as of 22:57, 31 January 2020
Contents
Problem 5
What is the sum of all real numbers 
for which 
Solution 1
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
The first case yields 
, which is equal to 
. Therefore, the two values for the positive case is 
and 
.
Similarly, taking the nonpositive case for the value inside the absolute value notation yields 
. Factoring and simplifying gives 
, so the only value for this case is 
.
Summing all the values results in 
.
Solution 2
We have the equations 
and 
.
Notice that the second is a perfect square with a double root at 
, and the first has real roots. By Vieta's, the sum of the roots of the first equation is 
. 
.
See Also
| 2020 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
