Difference between revisions of "1950 AHSME Problems/Problem 24"

Revision as of 17:46, 9 July 2015

Problem

The equation $x + \sqrt{x-2} = 4$ has:

$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$

Solution

$x + \sqrt{x-2} = 4$ Original Equation

$\sqrt{x-2} = 4 - x$ Subtract x from both sides

$x-2 = 16 - 8x + x^2$ Square both sides

$x^2 - 9x + 18 = 0$ Get all terms on one side

$(x-6)(x-3) = 0$ Factor

$x = \{6, 3\}$

If you put down A as your answer, it's wrong. You need to check for extraneous roots.

$6 + \sqrt{6 - 2} = 6 + \sqrt{4} = 6 + 2 = 8 \ne 4$

$3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark$

There is $\boxed{\textbf{(E)} \text{1 real root}}$

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AHSME Problems and Solutions

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

@@ Line 24: / Line 24: @@
 <math>3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark</math>
-There is <math>\textbf{(E)} \text{1 real root}</math>
+There is <math>\boxed{\textbf{(E)} \text{1 real root}}</math>
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Difference between revisions of "1950 AHSME Problems/Problem 24"

Revision as of 17:46, 9 July 2015

Problem

Solution

See Also