L=1/(square root of...)"8-2x-x^2"

Question

HELP!!!!!!!! I dont know where to start. solve for x

Answer 1

I'm not sure if you typed this in correctly. What I see is

 

L = 1/sqrt(8 - 2x-x²)

 

What is L?

 

Even if we don't know what L is, let's consider the right hand side (the function) first. If the function is supposed to equal 0 (a.k.a. L=0), the numerator has to be equal to 0, and the denominator can't.

 

0/1 = 0

 

But anything/0 ≠ 0, instead, it is equal to ± ∞, depending on the sign of the numerator.

 

Now, if L is some number other than 0, we do something a little different.

 

Multiply both sides by the square root term to get everything on one line.

 

L√(8 - 2x - x²) = 1

 

We want to solve for x, so let's get x alone, which means we should divide by L

 

√(8 - 2x - x²) = 1/L

 

Now square both sides

 

8 - 2x - x² = 1/L²

 

Put this in standard form by putting all the terms to one side of the equation. I like having my x² term positive, so I would put everything on the right side.

 

0 = x² + 2x + (1/L² - 8)

 

Now use the quadratic formula to solve for x.

 

x = [ -b±√(b²-4ac) ] / 2a

 

Since we don't know what L is, it will look very clumsy typed in here. I hope this helps!