1959 AHSME Problems/Problem 35

Problem 35

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: $\textbf{(A)}\ x\ge 0 \qquad\textbf{(B)}\ x\le n\qquad\textbf{(C)}\ x=0\qquad\textbf{(D)}\ \text{the set of all real numbers}\qquad\textbf{(E)}\ \text{none of these}$

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 $(E)$ none of these.

1959 AHSC (Problems • Answer Key • Resources)
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1959 AHSME Problems/Problem 35

Problem 35

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