Question

Anyone know the answer? Much appreciate the help!!

Answer 1

`x=3+\frac{\sqrt{4L^(2)+1}}{L}` and
`x=3-\frac{\sqrt{4L^(2)+1}}{L}`

Explanation:

we have

`L=\frac{1} {√(x^2-6x + 5)}`

Solve for x

That means-----> isolate the variable x

squared both sides

`L^(2) =(\frac{1} {√(x^2-6x + 5)})^(2)\\ \\L^(2)=\frac{1} {{x^2-6x + 5}}\\ \\x^2-6x + 5=(1)/(L^(2))\\ \\x^2-6x=(1)/(L^(2))-5\\ \\x^2-6x+9=(1)/(L^(2))-5 +9\\ \\x^2-6x+9=(1)/(L^(2))+4\\ \\(x-3)^2=(4L^(2)+1)/(L^(2))`

Take the square root both sides

`(x-3)=(+/-)\sqrt{(4L^(2)+1)/(L^(2))}\\ \\x=3(+/-)\frac{\sqrt{4L^(2)+1}}{L}`

therefore

`x=3+\frac{\sqrt{4L^(2)+1}}{L}`

`x=3-\frac{\sqrt{4L^(2)+1}}{L}`