Question about Casio FX991ES Scientific Calculator

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If the quadratic equation has no roots, you cannot find the roots. The discriminate is negative, so if we attempt to use the quadratic equation, we get no roots.

For example, y=x^2 +0x + 3

a=1, b=0, c=3

(-b+/- sqrt(b^2 -4ac))/2a

Substituting in the numbers, we get

x= (-0 +/-sqrt(0^2 - 4(1)3))/2

x= (0 +/- sqrt (-12))/2

We cannot do the square root of -12. Therefore, there are no roots. This is the same as having no x-intercepts. The discriminant is b^2-4ac. In this case it is -12. Thus, there are no real roots.

However, you can still determine the maximum or minimum, the vertex, the axis of symmetry, the y-intercept and the stretch/compression. With this information you can graph the equation.

In this case, the y-intercept is 3, the vertex is at (0,3), the axis of symmetry is x=0, the minimum value of the function is 3.

Good luck.

Let me know if you have any questions.

Paul

Posted on Nov 05, 2015

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Posted on Jan 02, 2017

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The calculator cannot do symbolic algebra. If equation is aX^2+bX+c=0, write it in the form a(X^2+(b/a)X+(c/a))=0 and solve the quadratic equation X^2+(b/a)X+(c/a)=0. Get the approximate roots X1, and X2 (if they exist) and write your original equation in the for

a(X-X1)(X-X2)=0

Your quadratic polynomial is factored as** a(X-X1)(X-X2)**

a(X-X1)(X-X2)=0

Your quadratic polynomial is factored as

May 28, 2014 | Casio Office Equipment & Supplies

Use the quadratic formula, or factor the quadratic polynomial. Once factored into a product of two first degree binomials, the roots are obtained by setting (in TURN) each binomial factor equal to zero.

May 02, 2014 | Casio fx-300ES Calculator

Your scientific calculator is unable to solve complex equation with complex coefficients. You should try to solve by hand directly using the quadratic formula or by factoring the polynomial in z

failing that, another way would be to set z=x+iy, substitute this for z, carry out the algebra and try to separate real and imaginary parts. But your two equations will constitute a system of two quadratic equations. I am not aware of any general method to solve coupled nonlinear equations.

Good luck.

failing that, another way would be to set z=x+iy, substitute this for z, carry out the algebra and try to separate real and imaginary parts. But your two equations will constitute a system of two quadratic equations. I am not aware of any general method to solve coupled nonlinear equations.

Good luck.

Dec 20, 2012 | Casio FX-115ES Scientific Calculator

Press MODE 5 to bring up the equation types. Press 3 for the quadratic. Enter the three coefficients. Press = to see the first root. If the display shows x1 then press down to see the other other root. If the display only shows x then that's a double root. Given the two roots x1 and x2, the factors of the quadratic are (x-x1) and (x-x2).

Refer to "Equation Calculations" on page E-55 of the manual and examples #094 and #095 in the appendix for examples of solving quadratic equations. If the you need the manual and/or appendix, you can download them from

http://support.casio.com/manualfile.php?rgn=1&cid=004001004

Refer to "Equation Calculations" on page E-55 of the manual and examples #094 and #095 in the appendix for examples of solving quadratic equations. If the you need the manual and/or appendix, you can download them from

http://support.casio.com/manualfile.php?rgn=1&cid=004001004

May 11, 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

The form of quadratic equation is

The roots of the above quadratic equation are

The roots of the above quadratic equation are

Nov 07, 2010 | Computers & Internet

Hello,

The Casio FX-300ES does not handle symbolic algebra. So it cannot factor a general polynomial expression. The methods can be found in any book on Algebra.

However if you are interested in approximate factorization of quadratic and cubic polynomials, you can use the calculator to do that. It can solve aX^3 +bX^2+cX+d =0 and the quadratic equations.

If you want to factor a cubic polynomial P3(X) = aX^3+bX^2+cX+d , you write the corresponding cubic equation as**aX^3+bX^2+cX=d =0** , then you divide all terms of the equation by** a** to obtain

**X^3+(b/a)X^2+(c/a)X+(d/a)=0.**

You use the calculator to solve (approximately) this equation.

Suppose you find the 3 roots**X1,X2,and X3.** Then the polynomial X^3+(b/a)X^2+(c/a)X+(d/a) can be cast in the factored form (X-X1)(X-X2)(X-X3) and the original polynomial P3(X) can be written as

**P3(X) = a*(X-X1)(X-X2)(X-X3) **

You can handle the quadratic polynomial the same way.

P2(X) =a*(X-X1)(X-X2) where X1, X2 are the two real roots

Hope it helps.

The Casio FX-300ES does not handle symbolic algebra. So it cannot factor a general polynomial expression. The methods can be found in any book on Algebra.

However if you are interested in approximate factorization of quadratic and cubic polynomials, you can use the calculator to do that. It can solve aX^3 +bX^2+cX+d =0 and the quadratic equations.

If you want to factor a cubic polynomial P3(X) = aX^3+bX^2+cX+d , you write the corresponding cubic equation as

You use the calculator to solve (approximately) this equation.

Suppose you find the 3 roots

You can handle the quadratic polynomial the same way.

P2(X) =a*(X-X1)(X-X2) where X1, X2 are the two real roots

Hope it helps.

Sep 27, 2009 | Casio fx-300ES Calculator

Hello,

This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.

Let the equation be ax^2 +bx+c=0

You compute the discriminant Call it disc.

**disc= (b^2-4*a*c)**

1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.

Then** x1=(-b+square root of disc)/(2*a)**

and**x2= (-b-square root of disc)/(2*a)**

2.** If disc=0** , square root of disc =0 and** x1=x2=-b/(2*a)**

3.**If disc is negative, there are no real solutions.**

To find the solutions you**replace a, b, and c **by their numerical values, to calculate disc, its square root, and calculate X1 and X2.

Hope it helps.

This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.

Let the equation be ax^2 +bx+c=0

You compute the discriminant Call it disc.

1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.

Then

and

2.

3.

To find the solutions you

Hope it helps.

Jun 26, 2009 | Sharp EL-531VB Calculator

Hello,

This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.

Let the equation be ax^2 +bx+c=0

You compute the discriminant Callit disc.

**disc= (b^2-4*a*c)**

1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.

Then** x1=(-b+square root of disc)/(2*a)**

and**x2= (-b-square root of disc)/(2*a)**

2.** If disc=0** , square root of disc =0 and** x1=x2=-b/(2*a)**

3.**If disc is negative, there are no real solutions.**

To find the solutions you**replace a, b, and c **by their numerical values, to calculate disc, its square root, and calculate X1 and X2.

Hope it helps.

This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.

Let the equation be ax^2 +bx+c=0

You compute the discriminant Callit disc.

1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.

Then

and

2.

3.

To find the solutions you

Hope it helps.

May 17, 2009 | Sharp EL-531VB Calculator

Hello,

This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.

Let the equation be ax^2 +bx+c=0

You compute the discriminant Call it disc.

**disc= (b^2-4*a*c)**

1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.

Then** x1=(-b+square root of disc)/(2*a)**

and**x2= (-b-square root of disc)/(2*a)**

2.** If disc=0** , square root of disc =0 and** x1=x2=-b/(2*a)**

3.**If disc is negative, there are no real solutions.**

To find the solutions you**replace a, b, and c **by their numerical values, to calculate disc, its square root, and calculate X1 and X2.

Hope it helps.

This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.

Let the equation be ax^2 +bx+c=0

You compute the discriminant Call it disc.

1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.

Then

and

2.

3.

To find the solutions you

Hope it helps.

May 05, 2009 | Sharp EL-531VB Calculator

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