Question about Texas Instruments TI-89 Calculator

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Why the answer for d(x^3,x) is a real number not a equation?

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~

Posted by Anonymous on

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why the answer for d(x^3,x) is a real number not a - 305bb927-a7af-4b50-afd5-f9e69e2c903c.jpg

Posted on Sep 01, 2013

6 Suggested Answers

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Posted on Jan 02, 2017

MangostaRex
  • 6 Answers

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

  • 3 Answers

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

  • 7993 Answers

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.

56ccf66.jpg

Hope it helps.


Posted on Oct 18, 2009

  • 7993 Answers

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

  • 1 Answer

SOURCE: solving absolute value equations or inequalities not working

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

Posted on Sep 06, 2010

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Chacha answers


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.

Aug 17, 2014 | Computers & Internet

1 Answer

Chacha answers


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.

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

1 Answer

F(x) is cubic function with zeros at +3 and +2+2i, find equation for f(x)=-12 need help please


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)

Jun 24, 2014 | Office Equipment & Supplies

1 Answer

4x - 2x + 1 = 5 + x - 7


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.

May 17, 2012 | MathRescue Word Problems Of Algebra Lite

1 Answer

Edit Casio FX - 9860G SD have complex equations??


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.

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

1 Answer

Casio FX - 9860G SD have complex equations?? edit


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)

Mar 17, 2012 | Casio Office Equipment & Supplies

1 Answer

TI 89


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

Nov 25, 2011 | Texas Instruments TI-89 Calculator

1 Answer

5/4x-18=-3 Ti 84 plus One: just press enter/solve Two: press 2nd Test (Math) then press enter Every time I put in and equation using the above method for the equal sign. it either takes me back to...


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.)

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

1 Answer

In complex number mode, how to get out?


"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
4591a1b3e15019ae25ad5813d5744ca9.png
If coefficients are complex you should expect some complex roots. Right?
If the coefficients are REAL then depending on the discriminant
80efae5ccea2ac0702e1a9c3b25afd2b.png
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.

d64abc4.jpg










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

2 Answers

If there are 623 students and the ratio of girls is 2 to 3 boys. how many girls are there?


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.

Nov 19, 2009 | Office Equipment & Supplies

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