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Find all values of $x$ such that $\sqrt{4x^2} - \sqrt{x^2} = 6$.

AWESOMEEE  Aug 11, 2015

Best Answer 

 #2
avatar+18712 
+10

Find all values of $x$ such that $\sqrt{4x^2} - \sqrt{x^2} = 6$.

 

$$\small{\text{$
\begin{array}{rcl}
\sqrt{4x^2}-\sqrt{x^2} &=& 6\\
\sqrt{4}\sqrt{x^2}-\sqrt{x^2} &=& 6\\
2\sqrt{x^2}-\sqrt{x^2} &=& 6\\
\sqrt{x^2} &=& 6 \qquad | \qquad 1^2\\
x^2 &=& 36 \qquad | \qquad \pm\sqrt{}\\
x_{1,2} &=& \pm\sqrt{36} \\
x_{1,2} &=& \pm6 \\
\mathbf{x_1} & \mathbf{=} & \mathbf{6}\\
\mathbf{x_2} & \mathbf{=} & \mathbf{-6}
\end{array}
$}}$$

heureka  Aug 12, 2015
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2+0 Answers

 #1
avatar+26326 
+10

$$\sqrt{4x^2}-\sqrt{x^2}\rightarrow2|x|-|x|\rightarrow|x|$$

 

|x| = 6 when x = 6 and when x = -6

 

(Perhaps I should note that |x| means the absolute value of x)

.

Alan  Aug 11, 2015
 #2
avatar+18712 
+10
Best Answer

Find all values of $x$ such that $\sqrt{4x^2} - \sqrt{x^2} = 6$.

 

$$\small{\text{$
\begin{array}{rcl}
\sqrt{4x^2}-\sqrt{x^2} &=& 6\\
\sqrt{4}\sqrt{x^2}-\sqrt{x^2} &=& 6\\
2\sqrt{x^2}-\sqrt{x^2} &=& 6\\
\sqrt{x^2} &=& 6 \qquad | \qquad 1^2\\
x^2 &=& 36 \qquad | \qquad \pm\sqrt{}\\
x_{1,2} &=& \pm\sqrt{36} \\
x_{1,2} &=& \pm6 \\
\mathbf{x_1} & \mathbf{=} & \mathbf{6}\\
\mathbf{x_2} & \mathbf{=} & \mathbf{-6}
\end{array}
$}}$$

heureka  Aug 12, 2015

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