Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1955 AHSME Problems/Problem 12"

(Created page with "==Problem== The solution of <math>\sqrt{5x-1}+\sqrt{x-1}=2</math> is: <math>\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x...")
 
(→‎Solution)
Line 4: Line 4:
 
<math>\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0</math>
 
<math>\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0</math>
  
−
==Solution==
+
==Solution 1==
−
First, square both sides.
+
First, square both sides. This gives us
  
−
<math>\sqrt{5x - 1}^2</math> + <math>2</math> <math>\cdot</math> <math>\sqrt{5x - 1}</math> <math>\cdot</math> <math>\sqrt{x - 1}</math> + <math>\sqrt{x - 1}^2</math> = <math>4</math> <math>\Longrightarrow</math> <math>5x - 1</math> + <math>2</math> <math>\cdot</math> <math>\sqrt{(5x - 1) \cdot (x - 1)}</math> + <math>x - 1</math> = <math>4</math> <math>\Longrightarrow</math> <math>2</math> <math>\cdot</math> <math>\sqrt{5x^2 - 6x + 1}</math> + <math>6x - 2</math> = <math>4</math>
+
<cmath>\sqrt{5x-1}^2+2\cdot\sqrt{5x-1}\cdot\sqrt{x-1}+\sqrt{x-1}^2=4 \Longrightarrow 5x-1+2\cdot\sqrt{(5x-1)\cdot(x-1)}+x-1=4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}+6x-2=4</cmath>
−
Then, add <math>-6x</math> to both sides.
+
Then, adding <math>-6x</math> to both sides gives us
  
−
<math>2</math> <math>\cdot</math> <math>\sqrt{5x^2 - 6x + 1}</math> + <math>6x - 2</math>
+
<cmath>2\cdot\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}-2 =-6x+4</cmath>
 +
 
 +
After that, adding <math>2</math> to both sides will give us
 +
 
 +
<cmath>2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x-2</cmath>

Revision as of 22:21, 9 July 2018

Problem

The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:

$\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0$

Solution 1

First, square both sides. This gives us

\[\sqrt{5x-1}^2+2\cdot\sqrt{5x-1}\cdot\sqrt{x-1}+\sqrt{x-1}^2=4 \Longrightarrow 5x-1+2\cdot\sqrt{(5x-1)\cdot(x-1)}+x-1=4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}+6x-2=4\] Then, adding $-6x$ to both sides gives us

\[2\cdot\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}-2 =-6x+4\]

After that, adding $2$ to both sides will give us

\[2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x-2\]

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_12&oldid=96078"