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Difference between revisions of "1959 AHSME Problems/Problem 35"

(Created page with "== Problem 35== The symbol <math>\ge</math> means "greater than or equal to"; the symbol <math>\le</math> means "less than or equal to". In the equation <math>(x-m)^2-(x-n)^2=...")
 
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== Problem 35==
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== Problem ==
 
The symbol <math>\ge</math> means "greater than or equal to"; the symbol <math>\le</math> means "less than or equal to".
 
The symbol <math>\ge</math> means "greater than or equal to"; the symbol <math>\le</math> means "less than or equal to".
 
In the equation <math>(x-m)^2-(x-n)^2=(m-n)^2; m</math> is a fixed positive number, and <math>n</math> is a fixed negative number. The set of values x satisfying the equation is:
 
In the equation <math>(x-m)^2-(x-n)^2=(m-n)^2; m</math> is a fixed positive number, and <math>n</math> is a fixed negative number. The set of values x satisfying the equation is:
 
<math>\textbf{(A)}\ x\ge 0 \qquad\textbf{(B)}\ x\le n\qquad\textbf{(C)}\ x=0\qquad\textbf{(D)}\ \text{the set of all real numbers}\qquad\textbf{(E)}\ \text{none of these}    </math>
 
<math>\textbf{(A)}\ x\ge 0 \qquad\textbf{(B)}\ x\le n\qquad\textbf{(C)}\ x=0\qquad\textbf{(D)}\ \text{the set of all real numbers}\qquad\textbf{(E)}\ \text{none of these}    </math>
  
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Applying the [[difference of squares]] technique on this problem, we can see that <cmath>(x-m)^2-(x-n)^2=((x-m)+(x-n)) \cdot ((x-m)-(x-n))</cmath>, so <cmath>(x-m)^2-(x-n)^2=((x-m)+(x-n)) \cdot ((x-m)-(x-n))=(m-n)^2.</cmath> Simplifying gives us<cmath>(2x-m-n)\cdot(n-m)=(m-n)^2.</cmath>Negating <math>n-m</math> creates:<cmath>-(2x-m-n)\cdot(m-n)=(m-n)^2.</cmath>Dividing by <math>m-n</math>, <cmath>-(2x-m-n)=m-n</cmath> <cmath>-2x+m+n=m-n</cmath> <cmath>-2x+n=-n</cmath> <cmath>-2x=-2n</cmath> <cmath>x=n</cmath>Lastly, since n is a fixed negative number, x must also be a fixed negative number, so <math>x<0</math>. Since this answer is not <math>A, B, C, </math> or <math> D</math>, the solution must be <math>(E)</math> none of these.
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== Solution ==
  
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==See also==
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Applying the [[difference of squares]] technique on this problem, we can see that <cmath>(x-m)^2-(x-n)^2=((x-m)+(x-n)) \cdot ((x-m)-(x-n)),</cmath> so <cmath>(x-m)^2-(x-n)^2=((x-m)+(x-n)) \cdot ((x-m)-(x-n))=(m-n)^2.</cmath> Simplifying gives us<cmath>(2x-m-n)\cdot(n-m)=(m-n)^2.</cmath>Negating <math>n-m</math> creates:<cmath>-(2x-m-n)\cdot(m-n)=(m-n)^2.</cmath>Dividing by <math>m-n</math>, <cmath>-(2x-m-n)=m-n</cmath> <cmath>-2x+m+n=m-n</cmath> <cmath>-2x+n=-n</cmath> <cmath>-2x=-2n</cmath> <cmath>x=n</cmath>Lastly, since <math>n</math> is a fixed negative number, <math>x</math> must also be a fixed negative number, so <math>x<0</math>. Since this answer is not <math>A, B, C, </math> or <math> D</math>, the solution must be <math>\fbox{\textbf{(E) }none of these}</math>.
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== See also ==
 
{{AHSME 50p box|year=1959|num-b=34|num-a=36}}
 
{{AHSME 50p box|year=1959|num-b=34|num-a=36}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:44, 21 July 2024

Problem

The symbol $\ge$ means "greater than or equal to"; the symbol $\le$ means "less than or equal to". In the equation $(x-m)^2-(x-n)^2=(m-n)^2; m$ is a fixed positive number, and $n$ is a fixed negative number. The set of values x satisfying the equation is: $\textbf{(A)}\ x\ge 0 \qquad\textbf{(B)}\ x\le n\qquad\textbf{(C)}\ x=0\qquad\textbf{(D)}\ \text{the set of all real numbers}\qquad\textbf{(E)}\ \text{none of these}$

Solution

Applying the difference of squares technique on this problem, we can see that \[(x-m)^2-(x-n)^2=((x-m)+(x-n)) \cdot ((x-m)-(x-n)),\] so \[(x-m)^2-(x-n)^2=((x-m)+(x-n)) \cdot ((x-m)-(x-n))=(m-n)^2.\] Simplifying gives us\[(2x-m-n)\cdot(n-m)=(m-n)^2.\]Negating $n-m$ creates:\[-(2x-m-n)\cdot(m-n)=(m-n)^2.\]Dividing by $m-n$, \[-(2x-m-n)=m-n\] \[-2x+m+n=m-n\] \[-2x+n=-n\] \[-2x=-2n\] \[x=n\]Lastly, since $n$ is a fixed negative number, $x$ must also be a fixed negative number, so $x<0$. Since this answer is not $A, B, C,$ or $D$, the solution must be $\fbox{\textbf{(E) }none of these}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 36
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All AHSME Problems and Solutions

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