Question about Texas Instruments TI-89 Calculator

When i type in d(x^3,x) or other equation likes this, i always get a number like 147, the calculator use 7 as x value to solute these kind of question. how can i get the right equation like 3x^2 rather than a number??? Thank You~

Hi,

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Posted on Jan 02, 2017

SOURCE: how to use ti 89

Just plug in one side of the equation and find the answer. Then plug in the other side and compare answers.

Posted on Feb 24, 2008

SOURCE: how to graph linear equation in two variables

I am not quite sure what you are asking...

(1) if asking for graphing y= -2/3x+2, then press [diamond] [F1: Y=] type in .2/3x +2, see graph by [diamond][graph]

(2) if asking for graphing y= -2/3x+a, where a=3x, then first archive a=3x in Home screen by [home], type 3x [sto] a, then go back to y= menu, and type -2/3x+a in Y1

(3) if asking for graphing an equation that is dependent on 2 variables at the same time, use x-y-z axis (aka, change mode to graphing 3-D by [mode]graph [5:3D]) then enter desired equation in y= menu using Z function in terms of x and y aka z(x,y)= ...)

Posted on Jul 23, 2008

SOURCE: How do I solve a polynomial equation:

Hello,

Sorry, but what you wrote is not an equation but a polynomial expression. You want to solve the equation x^4+5x^3-3x^2-43x-60 =0.

The solve( command, can only be used with real numbers.

The** solve(** is available through the CATALOG :
[2nd][CATALOG], scroll down till you reach the command. Highlight it
and press [ENTER]. The command echodes on main screen as **solve(** .

You
complete the command by entering the expression (not the equation), the
name of the variable you solve for, the initial guess , and { lower
limit, upper limit} between curly brackets, and the closing parenthesis.

Exemple:**solve (x^4+5x^3-3x^2-43x-60 ****, x,0 {-5,0} ) [ENTER]**

should give you the negative root,**solve (x^4+5x^3-3x^2-43x-60 ****, x,0 {0,5} ) [ENTER]**

should give you the positive root.

It
is implied that the expression is 0, so you should not
insert =0, otherwise you get an error. Here for the lower limit is -5 you
must use the change sign symbol (-) under the 3 key, not the regular
MINUS.

You may ask how I knew that there were two roots when the equation is a quartic? By first graphing it to have an idea about where the roots lie and how many there are. You should always do that to speed up the search.

There is another way to zoom in on the roots: by drawing the graph and using the tools accessible under the [2nd][CALC] menu, namely the option [2:Zero]

The resolution of the TI83/84 is not good enough for this function that grows too fast, but I am inserting a picture of the curve from another calculator with a much better resolution.

Hope it helps.

Posted on Oct 18, 2009

SOURCE: I have an equation: 6-2(5x-8) = 3x - 4. I need to

Hello,

I will not ask what you want to do with the equation I will
answer your question about the equal sign. Press [2nd][CATALOG] you
access the catalog of available commands. The equal sign is near the
bottom of the list. So it is faster to scroll up from the top, and keep
scrolling up till you reach the =. Select it and press [ENTER], it will
echoe at the place where the cursor was before you press
[2nd][CATALOG]. Complete your equation.

Hope it helps.

Posted on Oct 21, 2009

SOURCE: solving absolute value equations or inequalities not working

3/4 y 5=3/5y-1

Posted on Sep 06, 2010

It seems to me that you are trying to solve the quadratic equation

aX^2+bX+c=10 with a=-3, b=3, c=15 or**-3X^2+3X+15=0**.

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation**-3X^2+3X+15=0**..

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

**Solution1 =X_1=(-3-SQRT(189))/(-2*3)=(1+SQRT(21))/2**

**Solution2 =X_2=(-3+SQRT(189))/(-2*3)=(1-SQRT(21))/2** or about -1.791287847

Here SQRT stands for square root.

aX^2+bX+c=10 with a=-3, b=3, c=15 or

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

Here SQRT stands for square root.

Aug 17, 2014 | Computers & Internet

It seems to me that you are trying to solve the quadratic equation

aX^2+bX+c=10 with a=-3, b=3, c=15 or**-3X^2+3X+15=0**.

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation**-3X^2+3X+15=0**..

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

**Solution1 =X_1=(-3-SQRT(189))/(-2*3)=(1+SQRT(21))/2**

**Solution2 =X_2=(-3+SQRT(189))/(-2*3)=(1-SQRT(21))/2** or about -1.791287847

Here SQRT stands for square root.

aX^2+bX+c=10 with a=-3, b=3, c=15 or

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

Here SQRT stands for square root.

Aug 17, 2014 | SoftMath Algebrator - Algebra Homework...

A cubic polynomial with real coefficients has three roots (in the field of Complex numbers). One root is real, the other two are **complex conjugate.**

Roots are z=+3,z=2+2i, z=2-2i

F(z)=(z-3)(z-2-2i)(z-2+2i)

Roots are z=+3,z=2+2i, z=2-2i

F(z)=(z-3)(z-2-2i)(z-2+2i)

Jun 24, 2014 | Office Equipment & Supplies

Start by combining like terms on each side of the equation.

Remember, when working with variables the real numbers always come last in the equation(i.e. 5+x-7=x+5-7)

Once you have this completed move all variables to one side and all whole numbers to the other side. Solve from there.

Remember, when working with variables the real numbers always come last in the equation(i.e. 5+x-7=x+5-7)

Once you have this completed move all variables to one side and all whole numbers to the other side. Solve from there.

May 17, 2012 | MathRescue Word Problems Of Algebra Lite

Thhe Casio FX-9860G SD can solve a polynomial equation
of degree 2 or 3 with REAL coefficients. If the complex MODE is set to
REAL it will find the real roots. If the complex mode is set to** a+ib**, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Alternatively, after you create the system of 4 linear equations you can use the matrix utility to find Real(x), Im(x), Real(y) and Im(y) and recompose the x and y.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Alternatively, after you create the system of 4 linear equations you can use the matrix utility to find Real(x), Im(x), Real(y) and Im(y) and recompose the x and y.

Mar 17, 2012 | Casio FX-9860G Graphic Calculator

The Casio FX-9860G SD can solve a polynomial equation
of degree 2 or 3 with REAL coefficients. If the complex MODE is set to
REAL it will find the real roots. If the complex mode is set to** a+ib**, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Mar 17, 2012 | Casio Office Equipment & Supplies

Remove battery from calculator and put it again. Then type in HOME 2nd F1 and 2:NewProb and ENTER. Try to solve any equations now. It may be resolve problem!

Posted: Stevan Milanovic

Posted: Stevan Milanovic

Nov 25, 2011 | Texas Instruments TI-89 Calculator

If you wan to solve an equation using the SOLVE feature you enter it as follows (your example)

You should rewrite it as 5/4*X-18+3 (= 0). Implicitly, the calculator assumes that the right side is ZERO and you do not enter any right side nor an equal sign.

Type in

5/4 [*] [X,T,theta, n] -18 +3 [ALPHA] [ENTER] (SOLVE)

You should get an answer -780/47 or the decimal equivalent (depending on display mode.)

You should rewrite it as 5/4*X-18+3 (= 0). Implicitly, the calculator assumes that the right side is ZERO and you do not enter any right side nor an equal sign.

Type in

5/4 [*] [X,T,theta, n] -18 +3 [ALPHA] [ENTER] (SOLVE)

You should get an answer -780/47 or the decimal equivalent (depending on display mode.)

Feb 12, 2011 | Texas Instruments TI-84 Plus Calculator

"Is there anything that I can do about this to stop my answers being complex numbers when solving cubic equations?"

You are making an assumption that is unwarranted.Namely that a cubic equation must necessarily have 3 real roots. This is not the case.

Let your equation be

If coefficients are complex you should expect some complex roots. Right?

If the coefficients are REAL then depending on the discriminant

you can have three cases

DELTA positive : three distinct real roots

DELTA=0 , the equation has a multiple root and all roots are REAL

DELTA negative: the equation has ONE real root and 2 complex roots that are complex conjugate of each other.

It suffices to look at the Tartaglia/Cardano expressions of the roots to know that more often than not there is going to be complex roots.

You are making an assumption that is unwarranted.Namely that a cubic equation must necessarily have 3 real roots. This is not the case.

Let your equation be

If coefficients are complex you should expect some complex roots. Right?

If the coefficients are REAL then depending on the discriminant

you can have three cases

DELTA positive : three distinct real roots

DELTA=0 , the equation has a multiple root and all roots are REAL

DELTA negative: the equation has ONE real root and 2 complex roots that are complex conjugate of each other.

It suffices to look at the Tartaglia/Cardano expressions of the roots to know that more often than not there is going to be complex roots.

May 05, 2010 | Casio CFX-9850G Plus Calculator

The best way to solve this is to develop one or more equations and then solve for the unknown numbers.

I solved this problem twice using different sets of equations to make sure I was right, and here is what I found:

1) I used the equations x + y = 623, and x = (2/3)y. In these equations, x is the number of girls and y is the number of boys.

Substituting the second equation into the first, I get (2/3)y + y = 623.

Adding the left side together I get : (5/3)y = 623.

Dividing both sides by (5/3) I get 373.8.

This means that x (the number of girls) is 249.2.

2) For the second attempt, I developed the equation: 2x+3x=623. Here 2x is the number of girls, and 3x is the number of boys.

Adding the left side I get: 5x = 623.

Dividing both sides by 5 I get: x = 124.6.

This means that the number of girls (2x) is 249.2. Just like in the first method.

But since you can't have a 1/5 of a girl, the answer must be 249 girls.

I solved this problem twice using different sets of equations to make sure I was right, and here is what I found:

1) I used the equations x + y = 623, and x = (2/3)y. In these equations, x is the number of girls and y is the number of boys.

Substituting the second equation into the first, I get (2/3)y + y = 623.

Adding the left side together I get : (5/3)y = 623.

Dividing both sides by (5/3) I get 373.8.

This means that x (the number of girls) is 249.2.

2) For the second attempt, I developed the equation: 2x+3x=623. Here 2x is the number of girls, and 3x is the number of boys.

Adding the left side I get: 5x = 623.

Dividing both sides by 5 I get: x = 124.6.

This means that the number of girls (2x) is 249.2. Just like in the first method.

But since you can't have a 1/5 of a girl, the answer must be 249 girls.

Nov 19, 2009 | Office Equipment & Supplies

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