Difference between revisions of "1955 AHSME Problems/Problem 12"
Awesomechoco (talk | contribs) (→Solution 1) |
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the equation is false. | the equation is false. | ||
Therefore the answer is <math>\fbox{{\bf(D)} x=1}</math> | Therefore the answer is <math>\fbox{{\bf(D)} x=1}</math> | ||
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| + | Solution by awesomechoco | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | ==See Also== | ||
| + | == See Also == | ||
| + | {{AHSME box|year=1955|num-b=11|num-a=13}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 00:37, 10 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
![\[2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4+2 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x+6\]](http://latex.artofproblemsolving.com/b/c/f/bcf43cb2bf1c0702bec77e1e706f49a0d5d562c9.png)
Next, we divide both sides by 2 which gives us
![\[\frac{2\cdot\sqrt{5x^2-6x+1}-2+2}{2}=\frac{-6x+4+2}{2} \Longrightarrow \sqrt{5x^2-6x+1}=-3x+3\]](http://latex.artofproblemsolving.com/c/c/6/cc66950478d8efbf8c68e7f25f0d95050804c7c8.png)
Finally, solving the equation, we get
![\[5x^2-6x+1=(-3x+3)^2 \Longrightarrow 5x^2-6x+1=9x^2-18x+9\]](http://latex.artofproblemsolving.com/1/7/d/17da8d631601f88faccee533638bc929491f42d9.png)
![\[\Longrightarrow 5x^2-6x+1-(9x^2-18x+9)=9x^2-18x+9-(9x^2-18x+9)\]](http://latex.artofproblemsolving.com/e/d/0/ed07341f5abc49bbfdc3463f9c5a2770c3c9bcc7.png)
![\[\Longrightarrow -4x^2+12x-8=0 \Longrightarrow -4(x-1)(x-2)=0\]](http://latex.artofproblemsolving.com/8/3/d/83de176eadcfa60fe4f69dd9297cc9e0b7d70006.png)
![\[\Longrightarrow x-1=0 \text{or}\ x-2=0 \Longrightarrow x=1 \text{or}\ x=2\]](http://latex.artofproblemsolving.com/a/7/3/a731664c3231075bde73fd3356675719211ff21b.png)
Plugging 1 and 2 into the original equation, , we see that when 
![\[\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot1-1}+\sqrt{1-1}=2 \Longrightarrow \sqrt4+\sqrt0=2 \Longrightarrow 2+0=2 \Longrightarrow 2=2\]](http://latex.artofproblemsolving.com/5/a/1/5a160162b9d23784aa134839db41efaa407a012c.png)
the equation is true. On the other hand, we note that when 
![\[\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot2-1}+\sqrt{2-1}=2 \Longrightarrow \sqrt9+\sqrt1=2 \Longrightarrow 3+1=2 \Longrightarrow 4=0\]](http://latex.artofproblemsolving.com/4/a/8/4a8f6e033752a132521364143014945238a3769e.png)
the equation is false.
Therefore the answer is 
Solution by awesomechoco
Solution 2
See Also
See Also
| 1955 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
