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 21:57, 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 |
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