Question

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

Answer 1

`L=(1)/(√(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*√(8-2x-x^2)=(1)/(√(8-2x-x^2))*√(8-2x-x^2) \\ \\ L*√(8-2x-x^2)=1 \end{gathered}`

2. Divide both sides of the equation into L:

`\begin{gathered} (L*√(8-2x-x^2))/(L)=(1)/(L) \\ \\ √(8-2x-x^2)=(1)/(L) \end{gathered}`

3. Square both sides of the equation:

`\begin{gathered} (√(8-2x-x^2))^2=((1)/(L))^2 \\ \\ 8-2x-x^2=(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=(-b\pm√(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}`