Difference between revisions of "1955 AHSME Problems/Problem 12"
Awesomechoco (talk | contribs) (→Solution 1) |
Megaboy6679 (talk | contribs) |
||
| (5 intermediate revisions by 3 users not shown) | |||
| Line 30: | Line 30: | ||
<cmath>\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</cmath> | <cmath>\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</cmath> | ||
the equation is false. | the equation is false. | ||
| − | Therefore the answer is <math>\ | + | Therefore the answer is <math>\boxed{{\textbf{(D) }} x=1}</math>. |
| − | + | ~awesomechoco | |
==Solution 2== | ==Solution 2== | ||
| + | Let us test the answer choices, for it is in this case simpler and quicker. <math>x=0</math> and <math>x=2/3</math> obviously doesn't work, since square roots of negative numbers are not real. <math>x=2</math> doesn't work either, because the first term is already bigger than 2, and the second term cannot be negative. By the process of elimination, the answer is <math>\boxed{{\textbf{(D) }} x=1}</math>. | ||
| + | |||
| + | ~megaboy6679 | ||
| − | |||
== See Also == | == See Also == | ||
{{AHSME box|year=1955|num-b=11|num-a=13}} | {{AHSME box|year=1955|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 19:48, 31 January 2023
Contents
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 
.
~awesomechoco
Solution 2
Let us test the answer choices, for it is in this case simpler and quicker. 
and 
obviously doesn't work, since square roots of negative numbers are not real. doesn't work either, because the first term is already bigger than 2, and the second term cannot be negative. By the process of elimination, the answer is .
~megaboy6679
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. 
