1950 AHSME Problems/Problem 24
Problem
The equation has:
Solutions
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.
Solution 3
We can create symmetry in the equation: Let , then we have The two roots are .
Notice, that the first root is extraneous as the range for the square root function is always the non-negative numbers (remember, negative numbers in square roots give imaginary numbers - imaginary numbers in square roots don't give negative numbers); thus, the only real root for occurs for the second root; squaring both sides and solving for gives .
~Vndom
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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