High-order equations from 2nd to 6th degree, High-order equations from 2nd to 6th degree -2 – Casio FX-9750GII User Manual

Page 104

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background image

4-2

 K EQUA
 (SIML)

(3)

 CU@U AU @U

@UEUBU@U
DUCU@U FU

 (SOLV)

• Internal calculations are performed using a 15-digit mantissa, but results are displayed using

a 10-digit mantissa and a 2-digit exponent.

• Simultaneous linear equations are solved by inverting the matrix containing the coefficients

of the equations. For example, the following shows the solution (x, y, z) of a simultaneous

linear equation with three unknowns.

Because of this, precision is reduced as the value of the determinant approaches zero. Also,

simultaneous equations with three or more unknowns may take a very long time to solve.

• An error occurs if the calculator is unable to find a solution.

• After calculation is complete, you can press

(REPT), change coefficient values, and then

re-calculate.

2. High-order Equations from 2nd to 6th Degree

Your calculator can be used to solve high-order equations from 2nd to 6th degree.

• Quadratic Equation:

ax

2

+

bx

+

c

= 0 (

a

p 0)

• Cubic Equation:

ax

3

+

bx

2

+

cx

+

d

= 0 (

a

p 0)

• Quartic Equation:

ax

4

+

bx

3

+

cx

2

+

dx

+

e

= 0 (

a

p 0)

1. From the Main Menu, enter the EQUA mode.

2. Select the POLY (Polynomial) mode, and specify the degree of the equation.

You can specify a degree 2 to 6.

3. Sequentially input the coefficients.

• The cell that is currently selected for input is highlighted. Each time you input a coefficient,

the highlighting shifts in the sequence:

a

m

b

m

c

m …

• You can also input fractions and values assigned to variables as coefficients.

• You can cancel the value you are inputting for the current coefficient by pressing

) at

any time before you press

U to store the coefficient value. This returns to the coefficient

to what it was before you input anything. You can then input another value if you want.

–1

=

x
y
z

a

1

b

1

c

1

a

2

b

2

c

2

a

3

b

3

c

3

d

1

d

2

d

3

–1

=

x
y
z

a

1

b

1

c

1

a

2

b

2

c

2

a

3

b

3

c

3

d

1

d

2

d

3

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