Question about Texas Instruments TI-84 Plus Calculator

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Posted on Jan 02, 2017

Unfortunately this calculator does not display results in radical form. If you enter SQRT(147) you get 12.12435565...

What you are asking for is called**rationalizing** an irrational expression. Usually one tries to get rid of radicals that appear in the denominators. Here is an example.

Suppose you have 1/(c+SQRT(d)), and you want to get rid of the radical in the denominator. How to do it?

Recall the identity** (a-b)*(a+b)= a^2-b^2**. It is true for any **a** and **b **

**Rationalizing a denominator** (usual case)

Now consider (c+SQRT(d)). Multiply it by (c-SQRT(d))

[c+SQRT(d)]*[c-SQRT(d)]=c^2 -[SQRT(d)]^2=**c^2-d**

You see that there is no radical.

Now take**1/(c+SQRT(d))**. To get rid of the radical from the denominator multiply it by (c-SQRT(d)). But to leave the value of your expression ** 1/(c+SQRT(d))** unchanged, you must multiply both numerator and denominator by (c-SQRT(d)).

The numerator becomes 1*(c-SQRT(d))=**c-SQRT(d)**, and the denominator **(c^2-d)**.

Finally, the rationalized form of the expression**1/(c+SQRT(d))** is

[**c-SQRT(d)**]**/[c^2-d]**.

**Rationalizing a numerator**

Sometimes, people have a radical in the numerator that they want to get rid of and have it in the denominator. The procedure is the same

Example:**(c+SQRT(d))**, the denominator here is 1

**(c+SQRT(d))**=(c+SQRT(d))*[c-SQRT(d)]/[c-SQRT(d)]. This gives

**[c^2-d]/[c-SQRT(d)]**

You have no radical in the numerator but there is one in the denominator. This is called rationalizing the numerator.

Your case is a lot simpler

SQRT(147)=7*SQRT(3)

Multiply it by SQRT(3)/SQRT(3) which is 1. This gives

SQRT(147)=7*3/SQRT(3)=21/SQRT(3)

**SQRT(147)= 21/SQRT(3)**

What you are asking for is called

Suppose you have 1/(c+SQRT(d)), and you want to get rid of the radical in the denominator. How to do it?

Recall the identity

Now consider (c+SQRT(d)). Multiply it by (c-SQRT(d))

[c+SQRT(d)]*[c-SQRT(d)]=c^2 -[SQRT(d)]^2=

You see that there is no radical.

Now take

The numerator becomes 1*(c-SQRT(d))=

Finally, the rationalized form of the expression

[

Sometimes, people have a radical in the numerator that they want to get rid of and have it in the denominator. The procedure is the same

Example:

You have no radical in the numerator but there is one in the denominator. This is called rationalizing the numerator.

Your case is a lot simpler

SQRT(147)=7*SQRT(3)

Multiply it by SQRT(3)/SQRT(3) which is 1. This gives

SQRT(147)=7*3/SQRT(3)=21/SQRT(3)

Feb 09, 2014 | Texas Instruments TI-30XA Calculator

You do not need a calculator for this proportion.

For a proportion a/b=c/d, the product of the extrems (a*d) is equal to the product of the means (b*c).

In your case x*x=8*8 or x^2=64.

The two solutions are x=+square root (64) or x=-square root(64).

For a less trivial equation, you can use the**solve** feature of the FX-119ES. To do so, use the ALPHA keyboard to enter your equation; once the equation is typed in, do not press the main equal sign, but the Solve key (the key is located near the top left corner of the keyboard). You will be prompted to solve for X?.

For a proportion a/b=c/d, the product of the extrems (a*d) is equal to the product of the means (b*c).

In your case x*x=8*8 or x^2=64.

The two solutions are x=+square root (64) or x=-square root(64).

For a less trivial equation, you can use the

Feb 24, 2012 | Casio FX-115ES Scientific Calculator

Hi mavekae,

There are lot of examples of rational equations on this website, kindly follow the link below:

http://www.tutor-homework.com/Math_Help/college_algebra/m3l2notes1.pdf

Hope it helps,

Thank you for using fixya.

There are lot of examples of rational equations on this website, kindly follow the link below:

http://www.tutor-homework.com/Math_Help/college_algebra/m3l2notes1.pdf

Hope it helps,

Thank you for using fixya.

Sep 26, 2011 | Office Equipment & Supplies

Press key Mode, first select option 5:EQN (by pressing key 5), then option 2: a_nX+b_nY+c_nZ=d_n (by pressin 2).

Now input coefficients from equations in table (each column is for one variable), just type in number and press =, then use arrows to move to another cell.

Now input coefficients from equations in table (each column is for one variable), just type in number and press =, then use arrows to move to another cell.

Sep 02, 2011 | Casio FX-115ES Scientific Calculator

The TI-30XA does not offer any symbolic algebra or equation solving functions. You will need to do some work on paper. However, its not super difficult.

Solving quadratic equations (finding the roots) involve three steps:

You should exercise this simplification with different examples until you can do it in your sleep. The goal is to get a form x² + px + q = 0, where in the above example case, p = -1 and q = -2.

Now, use the memory places #0 and #1 to store p and q, as in:

-1 [STO] 0

-2 [STO] 1

Next, you should determine the discriminant D = p² - 4q of the equation using your TI-30XA calculator. Having put the two values into the memory comes in handy here:

[RCL] 0 [x²] [-] 4 [x] [RCL] 1 [=] [STO] 2

Note, we put D into memory place #2, we will need it again later.

Now check the displayed value D:

and you'll receive the original, normalized equation.

As an exercise for your understanding: you can save a few keystrokes by saving -p to memory #0 instead of p. Try to figure out why this helps.

Solving quadratic equations (finding the roots) involve three steps:

- Simplify the equation and bring into the normalized form x² + px + q = 0.
- Determine the discriminant D = p² - 4q and check if there are two, one or no solutions at all.
- Calculate the root(s), if any.

You should exercise this simplification with different examples until you can do it in your sleep. The goal is to get a form x² + px + q = 0, where in the above example case, p = -1 and q = -2.

Now, use the memory places #0 and #1 to store p and q, as in:

-1 [STO] 0

-2 [STO] 1

Next, you should determine the discriminant D = p² - 4q of the equation using your TI-30XA calculator. Having put the two values into the memory comes in handy here:

[RCL] 0 [x²] [-] 4 [x] [RCL] 1 [=] [STO] 2

Note, we put D into memory place #2, we will need it again later.

Now check the displayed value D:

- if D
- if D = 0, there is exactly one solution: -p / 2. Key in [RCL] 0 [+=-] [÷] 2 [=] and the displayed value is the only solution to the quadratic equation.
- if D > 0, there are exactly two solutions:

Key in [RCL] 0 [+=-] [+] [RCL] 2 [] [=] [÷] 2 [=] to get the first root, and

[RCL] 0 [+=-] [-] [RCL] 2 [] [=] [÷] 2 [=] for the second root.

and you'll receive the original, normalized equation.

As an exercise for your understanding: you can save a few keystrokes by saving -p to memory #0 instead of p. Try to figure out why this helps.

Feb 11, 2011 | Texas Instruments TI-30XA 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

Press the TABLE key (to the right of the GRAPH key), then press F1 to select "TABLE".

Mar 29, 2010 | Texas Instruments TI-85 Calculator

Your Solver is a numerical solver not a symbolic one. You have to supply an initial guess at a solution and the Solver finds it. If you suspect, or know, that there are others jut run the solver again supplying the initial guess for each additional solution.

One easy way to locate the roots of an equation is to draw the function (without =0) and see on the graph the approximate values of x where the function vanishes. These approximate values of the roots can be used as initial guesses in the Solver.

One easy way to locate the roots of an equation is to draw the function (without =0) and see on the graph the approximate values of x where the function vanishes. These approximate values of the roots can be used as initial guesses in the Solver.

Nov 20, 2008 | Casio FX-9750GPlus Calculator

I'm not sure how to do x4,

but from -5x3 and on yes. Mode-5-4, and insert coefficients.

Here the solution:

1. 1.927

2. -0.413

3. -2.514

but from -5x3 and on yes. Mode-5-4, and insert coefficients.

Here the solution:

1. 1.927

2. -0.413

3. -2.514

Nov 08, 2008 | Casio FX-115ES Scientific Calculator

You can enter any root by typing the number, then hitting MATH and 5, which brings up the root symbol with the x in front of it: x√. So the fifth-root would be 5 -> MATH -> 5 and then whatever number you want to get the fifth-root for: 5x√10 for example. As someone else had mentioned, you can also raise it to a rational power: 3^(1/3) which would be the same as the cubed root of 3, but you could also type: 3 -> MATH -> 5 -> 3 and get the same answer, but looking like this in your calculator: 3x√3. The option for 4 actually is a predefined cubed-root, and the one for option 5 there is the root symbol that can be used with any number before it to get any root you want. There are no parenthesis as you get when using the predefined square-root and cubed-root functions, though, so you may want to type them in yourself if entering a long string of operations in the calculator at one time to make sure the calculator doesn't include numbers under the root that you don't want it to. But I guess it depends on preference in terms of what method you choose between the rational exponents or the root symbol (and most would go with whichever seems easiest and quickest to enter), but you asked specifically how to get the cubed-root on the TI-83 Plus, so there's my best attempt at answering your question.

Aug 22, 2008 | Texas Instruments TI-83 Plus Calculator

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