Difference between revisions of "1950 AHSME Problems/Problem 24"
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The two roots are <math>\sqrt{x-2} = -2, 1</math>. | The two roots are <math>\sqrt{x-2} = -2, 1</math>. | ||
− | 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 <math>x</math> occurs for the second root; squaring both sides and solving for <math>x</math> gives <math>x=3 \Rightarrow \boxed{\textbf{(E)} \text{1 real root}}</math> | + | 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 <math>x</math> occurs for the second root; squaring both sides and solving for <math>x</math> gives <math>x=3 \Rightarrow \boxed{\textbf{(E)} \text{1 real root}}</math>. |
~Vndom | ~Vndom |
Latest revision as of 00:57, 12 October 2020
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 | |
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.