how to solve this equation: 5 - x^2 = e^x

[sam2x6]

I don't know how to solve this equation 5-x^2=e^x, other than graphing and getting the intersection. Is there an algebraic method to solve this equation?

Thanks for the help

 

[stapel]

I don't know how to solve this equation 5-x^2=e^x, other than graphing and getting the intersection. Is there an algebraic method to solve this equation?

I don't know how well graphing would do, unless you're referring to having technology (like Excel or a graphing calculator) find the decimal approximation. You'd need calculus (like Newton's Method) or numerical means to find the intersection value yourself.

 

[Deleted member 4993]

I don't know how to solve this equation 5-x^2=e^x, other than graphing and getting the intersection. Is there an algebraic method to solve this equation?

Thanks for the help

As far as I know, there is no closed form solution to the above equation.

Using Newton-Raphson iteration scheme, we get x = 1.306559

 

[Ishuda]

I don't know how to solve this equation 5-x^2=e^x, other than graphing and getting the intersection. Is there an algebraic method to solve this equation?

Thanks for the help

Since e^x + x² - 5 goes to positive infinity as x goes to both positive and negative infinity, there will be an even number [including zero] of solutions [in this case two solutions]. If x were large negative and x about minus the square root of 5, f would be about zero. So maybe a good starting point for one solution in a Newton-Raphson method would -2.2 (a little less than negative than -\(\displaystyle \sqrt{5}\)). For the other root, if x is small positive e^x would be about 3 so x about the square root of 2, say 1.4, could be used for the other starting point.

-2.21143775884204 and 1.2411427583996 are the two answers I get through a spread sheet program.