solve for x : a^x+x+c=0?

Answer 1

There is no way to solve this algebraically.

First of all, one equation and three variables makes it impossible to solve in the first place.

Secondly, the form this takes makes it impossible to solve algebraically. Intuition plays a big part... otherwise Newtons tedious method will suffice.

Answer 2

This equation cannot be solved with the functions that you are likely to have learned about in high school. However, if you have access to the Lambert W function, then it's easy. Consider the following:

a^x + x + c = 0

First, simplify the equation by letting y = x+c. So we have:

a^(y-c) + y = 0

Subtract y from both sides and multiply by by a^(-y):

a^(-c) = -ya^(-y)

Write in exponential form:

a^(-c) = -ye^(-y ln a)

Multiply both sides by ln a:

a^(-c) ln a = (-y ln a)e^(-y ln a)

Apply the Lambert W function to both sides:

W(a^(-c) ln a) = -y ln a

Divide by -ln a:

-W(a^(-c) ln a)/ln a = y

Switch back to x and subtract c from both sides:

x = -W(a^(-c) ln a)/ln a - c

And we are done.

N.B. -- the Lambert W function is multivalued on the interval [-1/e, 0), so this equation may have more than one solution in the real numbers (e.g. if a<1). Be sure to check both possible values of the Lambert W function, in this case.

Answer 3

This looks like transcendental equation. Taylor's formula or Newtons method probably would work.

Answer 4

ax^2 + x + c = 0

formula: x = ( -B +- SQRT( B2 - 4AC ) ) / 2A

here A=a , B=1, C=c

Therefore ,

x= ( -1 +- SQRT( 2 - 4ac ) ) / 2a

Answer 5

ax^2 + x + c = 0
then x = 1/2a [ -- 1 ±√{(--1)^2 -- 4ac}] = 1/2a[--1±√{1 -- 4ac}]