1956 AHSME Problems/Problem 42

Problem 42

The equation $\sqrt {x + 4} - \sqrt {x - 3} + 1 = 0$ has:

$\textbf{(A)}\ \text{no root} \qquad  \textbf{(B)}\ \text{one real root} \\ \textbf{(C)}\ \text{one real root and one imaginary root} \\ \textbf{(D)}\ \text{two imaginary roots} \qquad \qquad\textbf{(E)}\ \text{two real roots}$

Solution

Simplify. \[\sqrt {x + 4} = \sqrt{x-3} -1\] \[x + 4 = x - 3 -2\sqrt{x-3} + 1\] \[6 = -2\sqrt{x-3}\] We don't need to solve any further. The square root term is negative, therefore there will be no real roots. The answer is $\boxed{A}.$

-coolmath34

See Also

1956 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 41
Followed by
Problem 43
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