Difference between revisions of "2012 AMC 10B Problems/Problem 8"

m (Created page with "== Problem 8 == What is the sum of all integer solutions to <math>1<(x-2)^2<25</math>? <math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19...")
 
(Solutions)
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<math>(x-2)^2=4</math>  
 
<math>(x-2)^2=4</math>  
  
<math>x=4</math>
+
<math>x=4,0</math>
  
 
and
 
and
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<math>(x-2)^2=9</math>
 
<math>(x-2)^2=9</math>
  
<math>x=5</math>
+
<math>x=5,-1</math>
  
 
and
 
and
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<math>(x-2)^2=16</math>
 
<math>(x-2)^2=16</math>
  
<math>x=6</math>
+
<math>x=6,-2</math>
  
 
''What is the sum of all integer solutions''
 
''What is the sum of all integer solutions''
  
<math>4+5+6=\boxed{15}</math>
+
<math>4+5+6+0+(-1)+(-2)=\boxed{12}</math>
  
 
OR
 
OR
  
<math> \textbf{(C)}</math>
+
<math> \textbf{(B)}</math>

Revision as of 17:44, 24 February 2012

Problem 8

What is the sum of all integer solutions to $1<(x-2)^2<25$?

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\25$ (Error compiling LaTeX. ! Undefined control sequence.)

Solution



Solutions

$(x-2)^2$ = perfect square.

1< perfect square< 25

Perfect square can equal: 4, 9, or 16

Solve for x:

$(x-2)^2=4$

$x=4,0$

and

$(x-2)^2=9$

$x=5,-1$

and

$(x-2)^2=16$

$x=6,-2$

What is the sum of all integer solutions

$4+5+6+0+(-1)+(-2)=\boxed{12}$

OR

$\textbf{(B)}$

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