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Difference between revisions of "1950 AHSME Problems/Problem 24"
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There is <math>\textbf{(E)} \text{1 real root}</math>
 
There is <math>\textbf{(E)} \text{1 real root}</math>
  
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== See Also ==
 
{{AHSME box|year=1950|num-b=23|num-a=25}}
 
{{AHSME box|year=1950|num-b=23|num-a=25}}
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[[Category:Introductory Algebra Problems]]

Revision as of 13:20, 17 April 2012

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

See Also

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