Difference between revisions of "2020 AMC 10A Problems/Problem 5"
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<math>\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25</math> | <math>\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25</math> | ||
| − | == Solution == | + | == Solution 1== |
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive. | Split the equation into two cases, where the value inside the absolute value is positive and nonpositive. | ||
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Summing all the values results in <math>4+8+6=\boxed{\text{(C) }18}</math>. | Summing all the values results in <math>4+8+6=\boxed{\text{(C) }18}</math>. | ||
| + | |||
| + | == Solution 2== | ||
| + | We have the equations <math>x^2-12x+32=0</math> and x^2-12x+36=2<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}$. | ||
==See Also== | ==See Also== | ||
Revision as of 21:56, 31 January 2020
Contents
Problem 5
What is the sum of all real numbers 
for which 
Solution 1
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
The first case yields 
, which is equal to 
. Therefore, the two values for the positive case is 
and 
.
Similarly, taking the nonpositive case for the value inside the absolute value notation yields 
. Factoring and simplifying gives 
, so the only value for this case is 
.
Summing all the values results in 
.
Solution 2
We have the equations 
and x^2-12x+36=2$.
Notice that the second is a perfect square with a double root at$ (Error compiling LaTeX. Unknown error_msg)x=6
12
12+6=\boxed{\text{(C) }18}$.
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. 
