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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>
  
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== Solution ==  
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== 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>.
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== Solution 2==
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We have the equations <math>x^2-12x+32=0</math> and x^2-12x+36=2<math>.
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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

Problem 5

What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$

$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25$

Solution 1

Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.

The first case yields $x^2-12x+34=2$, which is equal to $(x-4)(x-8)=0$. Therefore, the two values for the positive case is $4$ and $8$.

Similarly, taking the nonpositive case for the value inside the absolute value notation yields $-x^2+12x-34=2$. Factoring and simplifying gives $(x-6)^2=0$, so the only value for this case is $6$.

Summing all the values results in $4+8+6=\boxed{\text{(C) }18}$.

Solution 2

We have the equations $x^2-12x+32=0$ 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$, and the first has real roots. By Vieta's, the sum of the roots of the first equation is$12$.$12+6=\boxed{\text{(C) }18}$.

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

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