Using The Quadratic Formula to Solve the System

Robert Argus, Aneesh Malhotra, Kathleen McLane

The second part of the project involved using the quadratic formula to solve the systems of equations. This was done by subtracting the last three equations from the first equation, to get three linear equations in four unknowns. These can be seen below:

\begin{align*} x(-2A_1+2A_2)+y(-2B_1+2B_2)+z(-2C_1+2C_2)+d(2c^2t_1-2c^2t_2)=A_1^2-A_2^2+B_1^2-B_2^2+C_1^2+C_2^2-c^2t_1^2+c^2t_2^2\\ x(-2A_1+2A_3)+y(-2B_1+2B_3)+z(-2C_1+2C_3)+d(2c^2t_1-2c^2t_3)=A_1^2-A_3^2+B_1^2-B_3^2+C_1^2+C_3^2-c^2t_1^2+c^2t_3^2\\ x(-2A_1+2A_4)+y(-2B_1+2B_4)+z(-2C_1+2C_4)+d(2c^2t_1-2c^2t_4)=A_1^2-A_4^2+B_1^2-B_4^2+C_1^2+C_4^2-c^2t_1^2+c^2t_4^2\\ \end{align*}

Using these equations, one equation could be created where x,y,z,d, and w are vectors. From this, formulas for x, y, and z can be solved in terms of d. These can be seen in the code. This equation can then be substitured into the first quadratic equation so that an equation is produced in one variable. The code for this work can be seen here G.m, with the code for the derivatives being seen here DG.m. The solution for this can be seen below.