How i can solve quadratic equations which have no real roots

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)

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.

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.

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

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.

The form of quadratic equation is


The roots of the above quadratic equation are

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.

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.

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.

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.