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

$2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x-2$

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1955 AHSME Problems/Problem 12

Problem

Solution 1