Numerical Solutions

You can find numerical solutions in two ways. You can apply Exact from the Solve submenu after entering at least one coefficient in floating-point form—that is, with a decimal. You can apply Numeric on the Solve submenu. Solve + Numeric gives all real solutions to a polynomial equation, but gives only one solution to a system of equations. You use it primarily when solving (systems of) transcendental equations or when you want to specify a search interval for the solution.

$\blacktriangleright$ Solve + Exact

x2 + 7x - 5.2 = 0, Solution is : $\left\{\vphantom{ x=-7.\,6773}\right.$x = - 7. 6773$\left.\vphantom{ x=-7.\,6773}\right\}$,$\left\{\vphantom{ x=.\,67732}\right.$x = . 67732$\left.\vphantom{ x=.\,67732}\right\}$


x3 -3. 8x - 15. 6 = 0, Solution is : $\left\{\vphantom{ x=3.0}\right.$x = 3.0$\left.\vphantom{ x=3.0}\right\}$,$\left\{\vphantom{ x=-1.\,5-1.\,7176i}\right.$x = - 1. 5 - 1. 7176i$\left.\vphantom{ x=-1.\,5-1.\,7176i}\right\}$,$\left\{\vphantom{ x=-1.\,5+1.\,7176i}\right.$x = - 1. 5 + 1. 7176i$\left.\vphantom{ x=-1.\,5+1.\,7176i}\right\}$

$\blacktriangleright$ Solve + Numeric

x2 + 7x - 5.2 = 0, Solution is : $\left\{\vphantom{ x=-7.\,6773}\right.$x = - 7. 6773$\left.\vphantom{ x=-7.\,6773}\right\}$,$\left\{\vphantom{ x=.\,67732}\right.$x = . 67732$\left.\vphantom{ x=.\,67732}\right\}$


x3 -3. 8x - 15. 6 = 0, Solution is : $\left\{\vphantom{ x=3.0}\right.$x = 3.0$\left.\vphantom{ x=3.0}\right\}$


$\left[\vphantom{ \begin{array}{c} x^{2}+y^{2}=5 \\ x^{2}-y^{2}=1 \end{array} }\right.$$\begin{array}{c} x^{2}+y^{2}=5 \\ x^{2}-y^{2}=1 \end{array}$$\left.\vphantom{ \begin{array}{c} x^{2}+y^{2}=5 \\ x^{2}-y^{2}=1 \end{array} }\right]$, Solution is : $\left\{\vphantom{ x=1.\,7321,y=-1.\,4142}\right.$x = 1. 7321, y = - 1. 4142$\left.\vphantom{ x=1.\,7321,y=-1.\,4142}\right\}$

$\blacktriangleright$ To find a numerical solution within a specified range of the variable

1.
Add a row to the bottom of the matrix, or press ENTER to generate a new input box in a display.

2.
Write the interval of your choice, and use the membership symbol ∈ to indicate that the variable lies in that interval.

$\blacktriangleright$ Solve + Numeric

x2 + y2 = 5
x2 - y2 = 1
x∈$\left(\vphantom{ -2,0}\right.$ -2, 0$\left.\vphantom{ -2,0}\right)$
, Solution is : $\left\{\vphantom{ x=-1.7321,y=-1.4142}\right.$x = - 1.7321, y = - 1.4142$\left.\vphantom{ x=-1.7321,y=-1.4142}\right\}$

x2 + y2 = 5
x2 - y2 = 1
x∈$\left(\vphantom{ -2,0}\right.$ -2, 0$\left.\vphantom{ -2,0}\right)$
y∈$\left(\vphantom{ 0,2}\right.$0, 2$\left.\vphantom{ 0,2}\right)$
, Solution is : $\left\{\vphantom{ y=1.4142,x=-1.7321}\right.$y = 1.4142, x = - 1.7321$\left.\vphantom{ y=1.4142,x=-1.7321}\right\}$

$\blacktriangleright$ To find all numerical solutions to a system of polynomial equations

1.
Change at least one of the coefficients to floating-point form.

2.
From the Solve submenu, choose Exact.

$\blacktriangleright$ Solve + Exact

x2 + y2 = 5.0
x2 - y2 = 1.0
, Solution is :
$\left\{\vphantom{ y=-1.4142,x=1.7321}\right.$y = - 1.4142, x = 1.7321$\left.\vphantom{ y=-1.4142,x=1.7321}\right\}$ 4pt
$\left\{\vphantom{ y=-1.4142,x=-1.7321}\right.$y = - 1.4142, x = - 1.7321$\left.\vphantom{ y=-1.4142,x=-1.7321}\right\}$ 4pt
$\left\{\vphantom{ y=1.4142,x=1.7321}\right.$y = 1.4142, x = 1.7321$\left.\vphantom{ y=1.4142,x=1.7321}\right\}$ 4pt
$\left\{\vphantom{ y=1.4142,x=-1.7321}\right.$y = 1.4142, x = - 1.7321$\left.\vphantom{ y=1.4142,x=-1.7321}\right\}$ 4pt

These four solutions are illustrated in the following graph as the four points of intersection of two curves. See the section Implicit PlotsDM6-4.tex#Implicit equations for guidelines on making such graphs.

$\blacktriangleright$ Plot 2D + Implicit

x2 + y2 = 5, x2 - y2 = 1

dtbpF3.0441in2.0384in0ptexample.wmf