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 $x$ is smaller than $3$ , $x + \sqrt{x-2} < 4$ , 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.

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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1950 AHSME Problems/Problem 24

Contents

Problem

Solution 1

Solution 2

See Also