Difference between revisions of "1950 AHSME Problems/Problem 24"

Latest revision as of 23:57, 11 October 2020

Problem

The equation $x + \sqrt{x-2} = 4$ has:

$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$

Solutions

Solution 1

$x + \sqrt{x-2} = 4$ 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 $\boxed{\textbf{(E)} \text{1 real root}}$

Solution 2

It's not hard to note that $x=3$ simply works, as $3 + \sqrt{1} = 4$ . But, $x$ is increasing, and $\sqrt{x-2}$ is increasing, so $3$ is the only root. If $x < 3$ , $x + \sqrt{x-2} < 4$ , and similarly if $x > 3$ , then $x + \sqrt{x-2} > 4$ . 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: $x+\sqrt{x-2} = 4$ $x-2+\sqrt{x-2} = 2.$ Let $y = \sqrt{x-2}$ , then we have $y^2+y-2 = 0$ $(y+2)(y-1) = 0$ The two roots are $\sqrt{x-2} = -2, 1$ .

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 $x$ occurs for the second root; squaring both sides and solving for $x$ gives $x=3 \Rightarrow \boxed{\textbf{(E)} \text{1 real root}}$ .

~Vndom

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.

@@ Line 1: / Line 1: @@
-== Problem==
+== Problem ==
 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
@@ Line 24: / Line 24: @@
 <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 ==
 {{AHSME 50p box|year=1950|num-b=23|num-a=25}}
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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