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# How do i enter a quadratic formula into this model calculator ?

Posted by Anonymous on

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I don't have the specific model of calculator that you referenced, but hope to give you guidance that will work on any calculator.

As an example, let's find the roots of 2x^2 - 10x + 12
In this case a=2 b=-10 and c=12.

(-(-10)+ sqrt ((-10)^2 -(4 * 2 * 12)))/(2*2)

and to find the other root,

(-(-10) - sqrt ((-10)^2 -(4 * 2 * 12)))/(2*2)

All the brackets may be necessary to make sure the calculator does the calculation correctly.

Take care,

Paul

Posted on Apr 04, 2015

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Posted on Jan 02, 2017

SOURCE: TI30Xa

Posted on Apr 11, 2008

I like to take the a, b, and c values and store them as variables  -  a->x b->y c->z
then run: (- y + SQRT ( y^2-(4xz) ) )/2x     AND  (- y - SQRT ( y^2-(4xz) ) )/2x

the parentheses may be a bit confusing - 1 holds the whole top of the formula to be divided together, one holds everything under the square root, one holds the -4ac (or -4xz).
That ought to cover it.

Posted on Jan 19, 2009

Hello,
What do you want to do with it? If you want just to enter it, you do so on the command line.
Here are exemples of solutions
Under [Mode][F2] exact/aprox set to exact
solve(x^2+x-5=0,x) finds the exact solutions the top ones

Under [MODE][F2] Exact/Aprox set to Approximate you get approximate numerical solutions, the bottom ones.
There is another command nSolve( which solves numerically for an equation. You can specify a range of values where solution is suspected to be.

Hope it helps

Posted on Sep 13, 2009

Hello,
The calculator does not have a solve program, but you can still use it to solve the quadratic equation with a little effort. If you know the therory skip toward the end.

Let aX^2+bX+C=0
1. First calculate the discriminant usually called Delta and given by
Delta =b^2 -4*a*c store the value in the variable D
If Delta is positive you have two roots X1 and X2 given by
X1=(-b+square root of Delta)/(2*a)
X2=(-b- square root of Delta)/(2*a)

If Delta is equal to 0, X1=X2=-b/(2*a)
If Delta i negative, no real solutions exist

You only solve if Delta is positive or equal to 0.

Putting in the values of a, b, and c
value of a (put you number) [SHIFT][STO] A
value of b (put your value) [SHIFT][STO] B
value of c (put you value) [SHIF][STO] C

[ALPHA] B [X^2] -4[x][ALPHA] A [x][ALPHA] C [=]
Value of delta is displayed. If it is positive, you store its square root in D

[Square root] [ANS] [SHIFT][STO] D ;

Calculate 1/(2*a) and store in variable F
1./(2[x] ALPHA A) [SHIFT][STO] F

To obtain X1
[ALPHA] F [x] ( (-) ALPHA B + ALPHA D ) [=]
To obtain X2
[ALPHA] F [x] ( (-) ALPHA B - ALPHA D ) [=]

Be careful: the (-) is the change sign not the regular minus sign.

Hope it helps.

Posted on Sep 24, 2009

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:

1. Simplify the equation and bring into the normalized form x² + px + q = 0.
2. Determine the discriminant D = p² - 4q and check if there are two, one or no solutions at all.
3. Calculate the root(s), if any.
Lets do (1) with an example, 3x - 2 = ( 6 + x ) / x.

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.
In the example, the discriminant turns out to be 9, which is greater than 0, so there are two different roots. Following the key sequences you should get the roots x1 = -1 and x2 = +2. You can double check by calculating

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.

Posted on Feb 12, 2011

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