1959 AHSME Problems/Problem 25

Problem 25

The symbol $|a|$ means $+a$ if $a$ is greater than or equal to zero, and $-a$ if a is less than or equal to zero; the symbol $<$ means "less than"; the symbol $>$ means "greater than." The set of values $x$ satisfying the inequality $|3-x|<4$ consists of all $x$ such that: $\textbf{(A)}\ x^2<49 \qquad\textbf{(B)}\ x^2>1 \qquad\textbf{(C)}\ 1<x^2<49\qquad\textbf{(D)}\ -1<x<7\qquad\textbf{(E)}\ -7<x<1$

Solution

The equation $|3-x| < 4$ can be solved by splitting it into two inequalities: $3-x<4$ and $3-x<-4$. The solutions to those inequalities are $x<-1$ and $x>7$, respectively. The common interval of those two inequalities is $\textbf{(D)}\ -1<x<7$

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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