Show that the equation 2x3−x2−3=02x^3 - x^2 - 3 = 02x3−x2−3=0 has a solution between x=1x = 1x=1 and x=2x = 2x=2.
Show that the equation 2x3−x2−3=02x^3 - x^2 - 3 = 02x3−x2−3=0 can be rearranged to give: x=32x−1\displaystyle x = \sqrt{\frac{3}{2x - 1}}x=2x−13
Starting with x0=1x_0 = 1x0=1, use the iteration formula xn+1=32xn−1\displaystyle x_{n+1} = \sqrt{\frac{3}{2x_n - 1}}xn+1=2xn−13 twice to find an estimate for the solution to 2x3−x2−3=02x^3 - x^2 - 3 = 02x3−x2−3=0
