Question about Texas Instruments TI-84 Plus Calculator

# Divisors of a number on the Ti-84 plus.

How do I find the divisor pair list on the Ti-84 plus? For example, 20 would be 1,20 or 4,5. I did this before using the y= function. y1 and y2 show the divisors on the table and x=20. It also involves using the the table set up and some other button. Thank you for helping me. :0)

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Hi there,
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Posted on Jan 02, 2017

SOURCE: table problem

I have had the same problem luckily i figured out what to do. 1. press the apps button 2. then there should be an unwanted app(it was the only one on mine) 3. pick it and hit enter 4. go to uninstall on the top of ur screen problem solved!!!! hope this helped

Posted on Oct 02, 2007

SOURCE: error :invalid DIM

One solution I know of is clearing the RAM. Sounds strange, doesn't it? But it fixed three different calcs.

Be sure to back up or archive important data before doing this.

Posted on Oct 21, 2008

Just hit the 2ND key and the 0 (catalog key)scroll down to percentage or just divide the answer by 100.

Posted on Feb 13, 2009

I am afraid you cannot use the TI8xPlus family of calculators to solve linear systems in matrix form. In this calculator, matrices must have real coefficients.
You can however separate (expand) the problem into a linear system of 4 equations in 4 unknowns and try to solve it with the calculator.

1. define X = a+i*b
2. Define Y=c+i*d
3. Rewrite the first equation substituting a+i*b for X and c+i*d for y.
4. Gather all real terms together, and all imaginary terms together on the left.
5. You will have (8a-15b-8c) +i(15a+8b-8d) = 10 +0i
6. This equation can be split into two equations by saying that
• the real part pf the left member is equal to the real part of the right member, this gives 8a-15b-8c=10, and
• the imaginary part of the left member is equal to the imaginary part of the right member, this gives 15a+8b-8d=0
Similarly, you rewrite the second complex equation substituting a+ib for X and c+id for Y, then expand the binomial products, gather the real parts on the left together, and the imaginary parts on the left together. After that you equate the real part on the left to the real part on the right, and do the same for the imaginary parts.

If I did not make mistakes during the expansions and the gathering of terms you should get the following equation
-(8a+8c+10d) +i*(-8b+10c-8d) =0+i*0 from which you extract an equation for the real parts, -(8a+8c+10d)=0 and another for the imaginagy parts i*(-8b+10c-8d) =i*0

If I did not make mistakes (you should be able to find them, if any) your system of two linear equations with complex coefficents has been converted to a system of 4 linear equations with real coefficients.

8a-15b-8c=10
15a+8b-8d=0
8a+8c+10d=0
-8b+10c-8d =0

Now, you can in theory solve this system with help of the calculator, to find a, b, c, and d. When these are found, you can reconstruct the X and Y solutions.

Now get to work: Ascertain that my extracted equations are correct, then solve for a, b,c, and d, and reconstruct X and Y.
I am no seer, but my hunch is that this system is degenarate. I will not explain what that means.

Posted on Feb 11, 2010

i would say use the normal key pad or send it in to T.I to fix.

Posted on Mar 12, 2010

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