How to root My Lava X1 - Vehicle Parts & Accessories

Hi rayalak857,

Have a nice day.


Calculator does not have a solve program, but you can use it to solve the quadratic equation with a little effort. If you know the theory skip toward the end.
Here is the example:

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. Goodluck

Thank you for using fixya.

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

Sorry, but no! it is not capable of factoring anything. It can however solve for the roots of a polynomial equation.Write your polynomial as P(X)= a(X² +(b/a)X+c/a) =0. Use the calculator to solve the polynomial equation X² +(b/a)X+c/a =0 and find the roots X1, and X2. You will then be able to write the original polynomial as P(X)=a(X-X1)(X-X2). Beware that by using the calculator, the values of the roots are approximate.

The short story is that this calculator does have a computer algebra system or CAS and thus cannot factor polynomials with arbitrary (unknown) coefficients or known coefficients.
However if the coefficients are given you can ,if you are willing to travel that way, factor approximately a polynomial P(x).
Basically, the idea is that any polynomial P(X) of degree n can be written in the factored form (X-x_1)(X-x_2)...(X-x_n), where x_1, x_2, x_3,...x_n are the roots (real or complex) of the equation P(X)=0.


The procedure ( for a 3rd degree polynomial) is as follows: (the fixYa site parser will remove the plus signs, so I am writing the whole word plus instead of the mathematical sign
If you want to factor a cubic polynomial P3(X) = aX^3 plus bX^2 plus cX plus d , you write the corresponding cubic equation as aX^3 plus bX^2 plus cX plus d =0 , then you divide all terms of the equation by a to obtain

X^3 plus (b/a)X^2 plus (c/a)X plus (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 plus (b/a)X^2 plus (c/a)X plus (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.

To find the various roots you must use the solve( application.

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.

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.

Hello,
Sorry, but you cannot use this calculator to factorize a general polynomial.
1. It does not know symbolic algebra.
2. It can only manipulate numbers.

However if you have polynomials of degree 2 or 3, with numerical coefficients (no letters) you can set [MODE] to equation and use the equation solver to find the real roots of 2nd degree or 3rd degree polynomials. Assuming that your polynomials have real roots (X1, X2) for the polynomial of degree 2, or (X1, X2, X3) for the polynomial of degree 3, then it is possible to write
P2(X) =a*(X-X1)(X-X2)
P3(X)= a(X-X1)(X-X2)(X-X3)
This is an approximate factorization, except if your calculator configured in MathIO, has been able to find exact roots (fractions and radicals)
where a is the coefficient of the highest degree monomial aX^2 +...
or aX^3 +....

But I have a hunch that this is not what you wanted to hear.

Good luck.

Hello,
Sorry, but you cannot use this calculator to factor a general polynomial.
1. It does not know symbolic algebra.
2. It can only manipulate numbers.

However if you have polynomials of degree 2 or 3, with numerical coefficients (no letters) you can set [MODE] to Equation and use the equation solver to find the real roots of 2nd degree or 3rd degree polynomials. Assuming that your polynomials have real roots (X1, X2) for the polynomial of degree 2, or (X1, X2, X3) for the polynomial of degree 3, then it is possible to write

P2(X) =a*(X-X1)(X-X2)
P3(X)= a(X-X1)(X-X2)(X-X3)

where a is the coefficient of the highest degree monomial aX^2 +...
or aX^3 +....

This is an approximate factorization, except if your calculator configured in MathIO, has been able to find exact roots (fractions and radicals)

While the [MODE][5:Equation] only handles quadratic and cubic equations, the [SHIFT][SOLVE=] solver finds the roots of arbitarry expressions (not limited to polynomials). In principle you can use it to find the roots of an expression. If it is a polynomial of dgree higher that 3 you can factor it (approximately).

But I have a hunch that this is not what you wanted to hear.

Hope it helps.