Solve for m
(Lm^2 )+(Rm)+(1/C)=0
This is just a quadratic equation, from you Algebra I days. So, use the quadratic formula:
m = (-R±√(R^2-4L/C))/2L
and pick the root that makes physical sense.
To solve for m in the equation (Lm^2)+(Rm)+(1/C) = 0, we can use the quadratic formula since we have a quadratic equation in terms of m.
The quadratic formula is given by:
m = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = L, b = R, and c = 1/C. Substituting these values into the quadratic formula, we get:
m = (-R ± √(R^2 - 4L(1/C))) / (2L)
Now we simplify further:
m = (-R ± √(R^2 - 4LC)) / (2L)
This gives us the two possible solutions for m. Note that the solutions may be real or complex, depending on the values of L, R, and C.
To solve the equation (Lm^2) + (Rm) + (1/C) = 0 for m, we can use the quadratic formula. The quadratic formula is given by:
m = (-b ± √(b^2 - 4ac)) / (2a)
Comparing the equation (Lm^2) + (Rm) + (1/C) = 0 to the general quadratic equation ax^2 + bx + c = 0, we can see that:
a = L
b = R
c = 1/C
Substituting these values into the quadratic formula, we get:
m = (-(R) ± √((R)^2 - 4(L)(1/C))) / (2(L))
Simplifying further,
m = (-R ± √((R)^2 - 4LC)) / (2L)
So, the solution for m is either (-R + √((R)^2 - 4LC)) / (2L) or (-R - √((R)^2 - 4LC)) / (2L), depending on the values of R, L, and C.