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1955 AHSME Problems/Problem 12

Revision as of 21:23, 9 July 2018 by Awesomechoco (talk | contribs) (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...")
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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

First, square both sides.

$\sqrt{5x - 1}^2$ + $2$ $\cdot$ $\sqrt{5x - 1}$ $\cdot$ $\sqrt{x - 1}$ + $\sqrt{x - 1}^2$ = $4$ $\Longrightarrow$ $5x - 1$ + $2$ $\cdot$ $\sqrt{(5x - 1) \cdot (x - 1)}$ + $x - 1$ = $4$ $\Longrightarrow$ $2$ $\cdot$ $\sqrt{5x^2 - 6x + 1}$ + $6x - 2$ = $4$ Then, add $-6x$ to both sides.

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

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