Question

A) Show that the equation 20-x^3-7x^2=0 can be rearranged to give x=20/x^2 - 7

Answer 1

Answer:

Step-by-step explanation:

      $20-x^3-7x^2=0$

a)

• Add $x^3$ to both sides of the equation:     $20-7x^2=x^3$
• Divide both sides by $x^2$:    $\frac{20}{x^2} -\frac{7x^2}{x^2} =\frac{x^3}{x^2}$  ⇒  $\frac{20}{x^2} -7 =x$

b)

    $x_1=\frac{20}{(x_0)^2} -7$

⇒ $x_1=\frac{20}{(-9)^2} -7$

⇒ $x_1=-\frac{547}{81}$

    $x_2=\frac{20}{(x_1)^2} -7$

⇒ $x_2=\frac{20}{(-\frac{547}{81})^2} -7$

⇒ $x_2=-6.561443673...$

   $x_3=\frac{20}{(x_2)^2} -7$

⇒ $x_3=\frac{20}{(-6.561443673...)^2} -7$

⇒ $x_3=-6.535451368...$

c) approximation to the location of one of the roots of the equation.  Each iteration gives a slightly more accurate value of a root $x$.