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

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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.
 
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.
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 +
==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>\sqrt{x-2}=y</math>, then we have
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<cmath>y^2+y-2=0</cmath>
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<cmath>(y+2)(y-1)=0</cmath>
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the two roots are <math>\sqrt{x-2}=-2, \sqrt{x-2}=1</math>
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 +
notice, that the first root is extraneous as the range for a square root is always the non-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>\boxed{x=3} \Rightarrow \text{(C) 1 real solution}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 00:58, 10 June 2020

Problem

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

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

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 $\sqrt{x-2}=y$, then we have \[y^2+y-2=0\] \[(y+2)(y-1)=0\] the two roots are $\sqrt{x-2}=-2, \sqrt{x-2}=1$

notice, that the first root is extraneous as the range for a square root is always the non-negative numbers; thus, the only real root for $x$ occurs for the second root; squaring both sides and solving for $x$ gives $\boxed{x=3} \Rightarrow \text{(C) 1 real solution}$

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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