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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...")
 
 
(9 intermediate revisions by 3 users not shown)
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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>
  
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==Solution==
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==Solution 1==
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First, square both sides.
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First, square both sides. This gives us
  
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<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>
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<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>
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Then, add <math>-6x</math> to both sides.
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Then, adding <math>-6x</math> to both sides gives us
  
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<math>2</math> <math>\cdot</math> <math>\sqrt{5x^2 - 6x + 1}</math> + <math>6x - 2</math>
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<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
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<cmath>2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4+2 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x+6</cmath>
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Next, we divide both sides by 2 which gives us
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<cmath>\frac{2\cdot\sqrt{5x^2-6x+1}-2+2}{2}=\frac{-6x+4+2}{2} \Longrightarrow \sqrt{5x^2-6x+1}=-3x+3</cmath>
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Finally, solving the equation, we get
 +
 
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<cmath>5x^2-6x+1=(-3x+3)^2 \Longrightarrow 5x^2-6x+1=9x^2-18x+9</cmath>
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<cmath>\Longrightarrow 5x^2-6x+1-(9x^2-18x+9)=9x^2-18x+9-(9x^2-18x+9) </cmath>
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<cmath>\Longrightarrow -4x^2+12x-8=0 \Longrightarrow -4(x-1)(x-2)=0</cmath>
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<cmath>\Longrightarrow x-1=0 \text{or}\ x-2=0 \Longrightarrow x=1 \text{or}\ x=2</cmath>
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Plugging 1 and 2 into the original equation, <math>\sqrt{5x-1}+\sqrt{x-1}=2</math>, we see that when <math>x=1</math>
 +
 
 +
<cmath>\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</cmath>
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the equation is true. On the other hand, we note that when <math>x=2</math>
 +
 
 +
<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>
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the equation is false.
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Therefore the answer is <math>\boxed{{\textbf{(D) }} x=1}</math>.
 +
 
 +
~awesomechoco
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==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 ==
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{{AHSME box|year=1955|num-b=11|num-a=13}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 20:48, 31 January 2023

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

\[2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4+2 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x+6\] Next, we divide both sides by 2 which gives us

\[\frac{2\cdot\sqrt{5x^2-6x+1}-2+2}{2}=\frac{-6x+4+2}{2} \Longrightarrow \sqrt{5x^2-6x+1}=-3x+3\] Finally, solving the equation, we get

\[5x^2-6x+1=(-3x+3)^2 \Longrightarrow 5x^2-6x+1=9x^2-18x+9\] \[\Longrightarrow 5x^2-6x+1-(9x^2-18x+9)=9x^2-18x+9-(9x^2-18x+9)\] \[\Longrightarrow -4x^2+12x-8=0 \Longrightarrow -4(x-1)(x-2)=0\] \[\Longrightarrow x-1=0 \text{or}\ x-2=0 \Longrightarrow x=1 \text{or}\ x=2\] Plugging 1 and 2 into the original equation, $\sqrt{5x-1}+\sqrt{x-1}=2$, we see that when $x=1$

\[\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot1-1}+\sqrt{1-1}=2 \Longrightarrow \sqrt4+\sqrt0=2 \Longrightarrow 2+0=2 \Longrightarrow 2=2\] the equation is true. On the other hand, we note that when $x=2$

\[\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot2-1}+\sqrt{2-1}=2 \Longrightarrow \sqrt9+\sqrt1=2 \Longrightarrow 3+1=2 \Longrightarrow 4=0\] the equation is false. Therefore the answer is $\boxed{{\textbf{(D) }} x=1}$.

~awesomechoco

Solution 2

Let us test the answer choices, for it is in this case simpler and quicker. $x=0$ and $x=2/3$ obviously doesn't work, since square roots of negative numbers are not real. $x=2$ 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 $\boxed{{\textbf{(D) }} x=1}$.

~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


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