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

## 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 1

$x + \sqrt{x-2} = 4$ 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 $\boxed{\textbf{(E)} \text{1 real root}}$

## Solution 2

It's not hard to note that $x=3$ simply works, as $3 + \sqrt{1} = 4$. But, $x$ is increasing, and $\sqrt{x-2}$ is increasing, so $3$ is the only root. If $x < 3$, $x + \sqrt{x-2} < 4$, and similarly if $x > 3$, then $x + \sqrt{x-2} > 4$. Imaginary roots don't have to be considered because squaring the equation gives a quadratic, and quadratics either have only imaginary or only real roots.

## See Also

 1950 AHSC (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 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.

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