Difference between revisions of "1950 AHSME Problems/Problem 24"
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− | == Problem== | + | == Problem == |
− | |||
The equation <math>x + \sqrt{x-2} = 4</math> has: | The equation <math>x + \sqrt{x-2} = 4</math> has: | ||
− | <math> \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} </math> | + | <math>\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}</math> |
− | ==Solution== | + | == Solutions == |
+ | === Solution 1 === | ||
<math>x + \sqrt{x-2} = 4</math> Original Equation | <math>x + \sqrt{x-2} = 4</math> Original Equation | ||
Line 24: | Line 24: | ||
<math>3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark</math> | <math>3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark</math> | ||
− | There is <math>\textbf{(E)} \text{1 real root}</math> | + | There is <math>\boxed{\textbf{(E)} \text{1 real root}}</math> |
+ | |||
+ | === 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 < 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. | ||
+ | |||
+ | === Solution 3 === | ||
+ | We can create symmetry in the equation: | ||
+ | <cmath>x+\sqrt{x-2} = 4</cmath> | ||
+ | <cmath>x-2+\sqrt{x-2} = 2.</cmath> | ||
+ | Let <math>y = \sqrt{x-2}</math>, then we have | ||
+ | <cmath>y^2+y-2 = 0</cmath> | ||
+ | <cmath>(y+2)(y-1) = 0</cmath> | ||
+ | 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>. | ||
+ | |||
+ | ~Vndom | ||
== See Also == | == See Also == | ||
− | {{AHSME box|year=1950|num-b=23|num-a=25}} | + | {{AHSME 50p box|year=1950|num-b=23|num-a=25}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
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 | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.