Texas Instruments TI-89 Manual Book page 521

For the
EXACT
portions that cannot be solved are returned as
an implicit equation or inequality.
Use the "|" operator to restrict the solution
interval and/or other variables that occur in the
equation or inequality. When you find a solution
in one interval, you can use the inequality
operators to exclude that interval from
subsequent searches.
is returned when no real solutions are
false
found. true
that any finite real value of
equation or inequality.
Since
solve()
you can use "and," "or," and "not" to combine
results from
other Boolean expressions.
Solutions might contain a unique new
undefined variable of the form @
an integer in the interval 1–255. Such variables
designate an arbitrary integer.
In real mode, fractional powers having odd
denominators denote only the real branch.
Otherwise, multiple branched expressions such
as fractional powers, logarithms, and inverse
trigonometric functions denote only the
principal branch. Consequently,
produces only solutions corresponding to that
one real or principal branch.
Note: See also
.
zeros()
solve(equation1 and equation2 [and ... ], {varOrGuess1,
varOrGuess2 [, ... ]})
Returns candidate real solutions to the
simultaneous algebraic equations, where
each
varOrGuess
want to solve for.
Optionally, you can specify an initial guess
for a variable. Each
form:
variable
– or –
=
variable
For example,
504 Appendix A: Functions and Instructions
setting of the
Exact/Approx
is returned if solve() can determine
satisfies the
var
always returns a Boolean result,
with each other or with
solve()
j with j being
n
solve()
cSolve() , cZeros() , nSolve()
Boolean expression
specifies a variable that you
must have the
varOrGuess
real or non-real number
is valid and so is
x x=3
exact(solve((xì a)e^(x)=ë xù
mode,
(xì a),x)) ¸
In Radian angle mode:
solve(tan(x)=1/x,x)|x>0 and x<1
¸
solve(x=x+1,x) ¸
solve(x=x,x) ¸
2xì 11 and solve(x^2ƒ9,x) ¸
In Radian angle mode:
solve(sin(x)=0,x) ¸ x = @n1ø p
solve(x^(1/3)=ë 1,x) ¸
solve(‡(x)=ë 2,x) ¸
solve(ë ‡(x)=ë 2,x) ¸
, and
solve(y=x^2ì 2 and
x+2y=ë 1,{x,y}) ¸
.
x
+ x = 0 or x = a
e
x =.860...
false
true
x  1 and x ƒ ë 3
x = ë 1
false
x = 4
x=1 and y=ë 1
or x=ë 3/2 and y=1/4