# Totally lost. How do I do these seemingly easy mathematical functions in Java?

• 05-07-2013, 12:12 AM
calculatingexistence
Totally lost. How do I do these seemingly easy mathematical functions in Java?
How do I do this in Java?

Code:

```Center points 1,1 & 2,2 (One and two lines below are the equation of two circles) (x-1)^2 + (y-1)^2 = 1 (x-2)^2 + (y-2)^2 = 1 Do all the operations of the above equations x^2 – 2x + 1 + y^2 – 2y + 1 = 1 x^2 – 2x + 1 + y^2 – 2y = 0 x^2 – 4x + 4 + y^2 – 4y + 4 = 1 x^2 – 4x + 7 + y^2 – 4y = 0 Multiply one of the equations by -1 -x^2 + 2x – 1 – y^2 + 2y = 0 Add the equation that was multiplied by -1  to the equation that wasn't multiplied by -1 -2x + 6 – 2y = 0 Put the above sum in terms of x -2x = 2y -6 x = -y + 3 Take the equation that was multiplied by -1 before it was multiplied by -1 and substitute x for what x was found to equal in the above equation which was -y + 3 (-y + 3)^2 – 2(-y + 3) + 1 + y^2 – 2y = 0 Do all the operations of the above equation y^2 – 6y + 9 + 2y – 6 + 1 + y^2 – 2y = 0 2y^2 – 6y + 4 = 0 Use the quadratic formula to solve for y ax^2 + bx + c = 0 a = 2 | b = -6 | c = 4 y = ( -b + sqrt ( b^2 – 4ac ) ) / 2a y = ( -b – sqrt ( b^2 – 4ac ) ) / 2a```
• 05-07-2013, 01:10 AM
jim829
Re: Totally lost. How do I do these seemingly easy mathematical functions in Java?
Native java has no ability to solve equations. So you need to either install a package for that purpose or use basic algebra and isolate one variable in terms of another. Then it's plug and chug.

Also note:
Multiplication is done by using *
Exponentiation is done by using Math.pow()
Square root is done by using Math.sqrt()

^ is bitwise exclusive or so don't use it for exponentiation

Regards,
Jim
• 05-07-2013, 02:22 AM
calculatingexistence
Re: Totally lost. How do I do these seemingly easy mathematical functions in Java?
Thanks Jim.