Answer:
[tex]x=3+\frac{\sqrt{4L^{2}+1}}{L}[/tex] and [tex]x=3-\frac{\sqrt{4L^{2}+1}}{L}[/tex]
Step-by-step explanation:
we have
[tex]L=\frac{1} {\sqrt{x^2-6x + 5}}[/tex]
Solve for x
That means-----> isolate the variable x
squared both sides
[tex]L^{2} =(\frac{1} {\sqrt{x^2-6x + 5}})^{2}\\ \\L^{2}=\frac{1} {{x^2-6x + 5}}\\ \\x^2-6x + 5=\frac{1}{L^{2}}\\ \\x^2-6x=\frac{1}{L^{2}}-5\\ \\x^2-6x+9=\frac{1}{L^{2}}-5 +9\\ \\x^2-6x+9=\frac{1}{L^{2}}+4\\ \\(x-3)^2=\frac{4L^{2}+1}{L^{2}}[/tex]
Take the square root both sides
[tex](x-3)=(+/-)\sqrt{\frac{4L^{2}+1}{L^{2}}}\\ \\x=3(+/-)\frac{\sqrt{4L^{2}+1}}{L} [/tex]
therefore
[tex]x=3+\frac{\sqrt{4L^{2}+1}}{L}[/tex]
[tex]x=3-\frac{\sqrt{4L^{2}+1}}{L}[/tex]
