# 1955 AHSME Problems/Problem 12

## 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 