1955 AHSME Problems/Problem 12: Difference between revisions
Awesomechoco (talk | contribs) 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..." |
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<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 | ||
< | <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, | Then, adding <math>-6x</math> to both sides gives us | ||
< | <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 
is:
Solution 1
First, square both sides. This gives us
![\[\sqrt{5x-1}^2+2\cdot\sqrt{5x-1}\cdot\sqrt{x-1}+\sqrt{x-1}^2=4 \Longrightarrow 5x-1+2\cdot\sqrt{(5x-1)\cdot(x-1)}+x-1=4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}+6x-2=4\]](http://latex.artofproblemsolving.com/a/b/1/ab1749754f7c014be691a4378b6f8ea71e28ec7a.png)
Then, adding 
to both sides gives us
![\[2\cdot\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}-2 =-6x+4\]](http://latex.artofproblemsolving.com/1/4/8/14841ef20eece00fb76268aea2a342035b06f9f2.png)
After that, adding 
to both sides will give us
![\[2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x-2\]](http://latex.artofproblemsolving.com/4/4/2/4422db5e631d9c30b27c56c7ca0c0e45e5690c5a.png)
