Question

I need help!! I looked at my notes & nothing.

Answer 1

$L=\frac{1}{\sqrt{8-2x-x^2}}$

To solve x;

1. Mulriply both sides of the equation by the denominator of the fraction in the right:

$\begin{gathered} L*\sqrt{8-2x-x^2}=\frac{1}{\sqrt{8-2x-x^2}}*\sqrt{8-2x-x^2} \\ \\ L*\sqrt{8-2x-x^2}=1 \end{gathered}$

2. Divide both sides of the equation into L:

$\begin{gathered} \frac{L*\sqrt{8-2x-x^2}}{L}=\frac{1}{L} \\ \\ \sqrt{8-2x-x^2}=\frac{1}{L} \end{gathered}$

3. Square both sides of the equation:

$\begin{gathered} (\sqrt{8-2x-x^2})^2=(\frac{1}{L})^2 \\ \\ 8-2x-x^2=\frac{1}{L^2} \end{gathered}$

4. Rewrite the term in the right with a negative exponent:

$8-2x-x^2=L^{-2}$

3. Rewrite the equation in the form ax^2+bx+c=0

$-x^2-2x+8-L^{-2}=0$

Use the quadratic formula to solve x:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$\begin{gathered} a=-1 \\ b=-2 \\ c=8-L^{-2} \end{gathered}$$\begin{gathered} x=\frac{-(-2)\pm\sqrt{(-2)^2-4(-1)(8-L^{-2})}}{2(-1)} \\ \\ x=\frac{2\pm\sqrt{4+4(8-L^{-2})}}{-2} \\ \\ x=-\frac{2\pm\sqrt{4+4(8-L^{-2})}}{2} \end{gathered}$Answer:$x=\frac{-2\pm\sqrt{4+4(8-L^{-2})}}{2}$