How do I solve this equation for x?

  • Thread starter p75213
  • Start date
In summary, the conversation discusses solving a quadratic equation involving the variables R and x. The first equation is simplified to a quadratic form, and the importance of checking solutions is emphasized.
  • #1
p75213
96
0
Can somebody point me in the right direction with this?

[tex]\begin{array}{l}
\sqrt {{R^2} + {{\left( {xL - {\textstyle{1 \over {xC}}}} \right)}^2}} = \sqrt 2 R \\
{R^2} + {\left( {xL - {\textstyle{1 \over {xC}}}} \right)^2} = 2{R^2} \\
xL - {\textstyle{1 \over {xC}}} = R\sqrt {2 - 1} \\
\end{array}[/tex]
 
Mathematics news on Phys.org
  • #2
I'm not sure why you have a square root of 2-1, but it is correct I guess.

From this point, multiply x on both sides and use the quadratic formula.
 
  • #3
Don't forget that:

1. [itex]xL - {\textstyle{1 \over {xC}}} = -R\sqrt {2 - 1} \ [/itex] (Note the minus sign for R) may yield a solution as well.

2. You should check all solutions you get in the original equation. Since your work involved squaring both sides of an equation, it is possible to have extraneous solutions.
 
  • #4
Multiply both sides by x and you get a quadratic equation for x.
 
  • #5


To solve this equation for x, we need to isolate the variable x on one side of the equation. We can do this by first simplifying the equation and then using algebraic operations to manipulate it.

1. Simplify the equation by expanding the squared terms on both sides:
R^2 + x^2L^2 - 2xL/C + 1/x^2C^2 = 2R^2

2. Combine like terms on the left side of the equation:
x^2L^2 - 2xL/C + 1/x^2C^2 = R^2

3. Move all terms with x to one side of the equation by subtracting R^2 from both sides:
x^2L^2 - 2xL/C + 1/x^2C^2 - R^2 = 0

4. Factor out x from the first two terms on the left side:
x(xL^2 - 2L/C) + 1/x^2C^2 - R^2 = 0

5. To simplify the equation further, we can multiply both sides by x^2C^2:
x(xL^2 - 2L/C) + 1 - R^2x^2C^2 = 0

6. Use the quadratic formula to solve for x:
x = (2L/C ± √(4L^2/C^2 - 4(1-R^2C^2)))/2

7. Simplify the expression inside the square root:
x = (2L/C ± √(4L^2/C^2 - 4 + 4R^2C^2))/2

8. Simplify further and solve for x:
x = (2L/C ± √(4L^2/C^2 + 4R^2C^2 - 4))/2
x = (2L/C ± 2√(L^2/C^2 + R^2C^2 - 1))/2
x = L/C ± √(L^2/C^2 + R^2C^2 - 1)

Therefore, the solutions for x are:
x = L/C + √(L^2/C^2 + R^2C^2 - 1)
x = L/C - √(L^2
 

1. How do I solve a basic one-variable equation for x?

Solving a basic one-variable equation for x involves isolating the variable on one side of the equation and performing inverse operations on both sides until x is the only term left. Remember to perform the same operation on both sides of the equation to maintain equality.

2. What are the common rules for solving equations?

The common rules for solving equations are: 1) perform the inverse operation to get rid of constants and coefficients; 2) follow the order of operations; 3) combine like terms; 4) use the distributive property when necessary; and 5) check your solution by plugging it back into the original equation.

3. How do I solve equations with fractions?

To solve equations with fractions, you can either clear the fractions by multiplying both sides of the equation by the common denominator or cross-multiply. Remember to simplify the fractions before solving the equation.

4. What do I do if there are variables on both sides of the equation?

If there are variables on both sides of the equation, you can either combine like terms on one side and move the other terms to the opposite side, or you can subtract the variable on one side from both sides and then perform inverse operations to isolate the variable.

5. How do I solve equations with absolute value?

To solve equations with absolute value, you need to consider both the positive and negative solutions. You can rewrite the equation as two separate equations, one with the absolute value and one without, and then solve for both solutions. Remember to check your solutions by plugging them back into the original equation.

Similar threads

Replies
2
Views
1K
  • General Math
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
3
Views
195
  • General Math
Replies
2
Views
803
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
Replies
2
Views
1K
  • General Math
Replies
13
Views
1K
Back
Top