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

1956 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 41	Followed by Problem 43
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1956 AHSME Problems/Problem 42

Problem 42

Solution

See Also