Difference between revisions of "1950 AHSME Problems/Problem 24"
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==Solution 2== | ==Solution 2== | ||
− | It's not hard to note that <math>x=3</math> simply works, as <math>3 + \sqrt{1} = 4</math>. But, <math>x</math> is increasing, and <math>\sqrt{x-2}</math> is increasing, so <math>3</math> is the only root. If <math>x< | + | It's not hard to note that <math>x=3</math> simply works, as <math>3 + \sqrt{1} = 4</math>. But, <math>x</math> is increasing, and <math>\sqrt{x-2}</math> is increasing, so <math>3</math> is the only root. If <math>x < 3</math>, <math>x + \sqrt{x-2} < 4</math>, and similarly if <math>x > 3</math>, then <math>x + \sqrt{x-2} > 4</math>. 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 == | == See Also == |
Revision as of 17:05, 11 May 2019
Contents
Problem
The equation has:
Solution 1
Original Equation
Subtract x from both sides
Square both sides
Get all terms on one side
Factor
If you put down A as your answer, it's wrong. You need to check for extraneous roots.
There is
Solution 2
It's not hard to note that simply works, as . But, is increasing, and is increasing, so is the only root. If , , and similarly if , then . 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 by Problem 23 |
Followed by Problem 25 | |
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All AHSME Problems and Solutions |
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