+0

+1
124
2

Solve for x: $$\large \dfrac{\color{red}{ \sqrt{2x-1} }+\color{green}{\sqrt{2x+1}} }{\color{green}{ \sqrt{2x+1}}-\color{red}{\sqrt{2x-1}} }=\color{blue}{\dfrac{5}{3}} \quad, \quad \color{magenta}{x} = \ ?$$

Jul 1, 2020

#1
+732
0

hi guest!

i honestly don't know how to do this, but if you need quick answers, the answer is $$\boxed{x=\frac{17}{30}}$$ (according to wolfram alpha)

ps how did you make the text colored? it looks really cool :)

Jul 1, 2020
#2
+25597
+3

Solve for x:
$$\large \dfrac{\color{red}{ \sqrt{2x-1} }+\color{green}{\sqrt{2x+1}} }{\color{green}{ \sqrt{2x+1}}-\color{red}{\sqrt{2x-1}} }=\color{blue}{\dfrac{5}{3}} \quad, \quad \color{magenta}{x} = \ ?$$

$$\begin{array}{|rcll|} \hline \dfrac{\color{red}{ \sqrt{2x-1} }+\color{green}{\sqrt{2x+1}} }{\color{green}{ \sqrt{2x+1}}-\color{red}{\sqrt{2x-1}} } &=& \color{blue}{\dfrac{5}{3}} \\\\ \dfrac{ \color{red}{ \Big(\sqrt{2x-1} }+\color{green}{\sqrt{2x+1}} \Big)} { \color{green}{ \Big( \sqrt{2x+1}}-\color{red}{\sqrt{2x-1}} \Big) } * \dfrac{ \Big( \sqrt{2x-1}+ \sqrt{2x+1} \Big)} { \Big( \sqrt{2x+1}+ \sqrt{2x-1} \Big) } &=& \color{blue}{\dfrac{5}{3}} \\\\ \dfrac{ \Big( \sqrt{2x-1}+ \sqrt{2x+1} \Big)^2} { 2x+1-(2x-1) } &=& \color{blue}{\dfrac{5}{3}} \\\\ \dfrac{ 2x-1 +2\sqrt{(2x-1)(2x+1)}+2x+1 } { 2x+1-(2x-1) } &=& \color{blue}{\dfrac{5}{3}} \\\\ \dfrac{ 4x +2\sqrt{(2x-1)(2x+1)} } { 2x+1-2x+1 } &=& \color{blue}{\dfrac{5}{3}} \\\\ \dfrac{ 4x +2\sqrt{(2x-1)(2x+1)} } { 2 } &=& \color{blue}{\dfrac{5}{3}} \\\\ \dfrac{ 4x +2\sqrt{4x^2-1} } { 2 } &=& \color{blue}{\dfrac{5}{3}} \\\\ 2x +\sqrt{4x^2-1} &=& \color{blue}{\dfrac{5}{3}} \\\\ \sqrt{4x^2-1} &=& \color{blue}{\dfrac{5}{3}} - 2x \quad | \quad \text{square both sides} \\\\ 4x^2-1 &=& \left( \color{blue}{\dfrac{5}{3}} - 2x\right)^2 \\\\ 4x^2-1 &=& \dfrac{25}{9}-\dfrac{20x}{3} + 4x^2 \\\\ -1 &=& \dfrac{25}{9}-\dfrac{20x}{3} \\\\ \dfrac{20x}{3} &=& \dfrac{25}{9} + 1 \\\\ \dfrac{20x}{3} &=& \dfrac{34}{9} \\\\ x &=& \dfrac{34}{9}* \dfrac{3}{20} \\\\ x &=& \dfrac{17}{3*10} \\\\ \mathbf{x} &=& \mathbf{\dfrac{17}{30}} \\ \hline \end{array}$$

Jul 2, 2020