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8 years ago

you need practise for it!!!

the other formula -b +- ....

is still available..

8 years ago

its very easy

first method is by discriminant ..whichtells us the nature of the roots,

and after that apply graphical method which tells us about their signs.

8 years ago

i think you know how to take out descrimnant. aftr you get it check the value of d if, d=o then it is real and equal or if d>0 then it is unequal and real and f d<0 then i is imaginary.

8 years ago

The 4 roots () for any quartic equation;

where are equal to those of

where , , and .

The roots in terms of these 4 coefficients are given by the formula in the image below:

However, this formula is too unwieldy for general use, hence other methods or simpler formulas (given below) are generally used.^{[7]}

Consider the quartic

If *a*_{0} = 0 then *Q*(0) = 0, and so *x* = 0 is a solution. It follows that *Q*(*x*) may be factorised as *Q*(x) = *x*·(*a*_{4}*x*^{3} + *a*_{3}*x*^{2} + *a*_{2}*x* + *a*_{1}). The remaining three roots (see Fundamental Theorem of Algebra) can be found by solving the cubic equation *a*_{4}*x*^{3} + *a*_{3}*x*^{2} + *a*_{2}*x* + *a*_{1} = 0.

If then , so is a root. Similarly, if that is, then is a root.

When is a root, we can divide by and get

where is a cubic polynomial, which may be solved to find ''s other roots. Similarly, if is a root,

where is some cubic polynomial.

If then −*k* is a root and we can factor out ,

And if then both and are roots Now we can factor out and get

To get *Q* ''s other roots, we simply solve the quadratic factor.

If then

We call such a polynomial a **biquadratic**, which is easy to solve.

Let Then *Q* becomes a quadratic *q* in

Let and be the roots of *q*. Then the roots of our quartic *Q* are

Steps:

- Divide by
*x*^{ 2}. - Use variable change
*z*=*x*+*m*/*x*.

To begin, the quartic must first be converted to a *depressed quartic*.

Let

be the general quartic equation we want to solve. Divide both sides by *A* to produce a monic polynomial,

The first step should be to eliminate the *x*^{3} term. To do this, change variables from *x* to *u*, such that

- .

Then

Expanding the powers of the binomials produces

Collecting the same powers of *u* yields

Now rename the coefficients of *u*. Let

The resulting equation is

which is a **depressed quartic equation**.

If then we have a biquadratic equation, which (as explained above) is easily solved; using reverse substitution we can find our values for .

If then one of the roots is and the other roots can be found by dividing by , and solving the resulting depressed cubic equation,

Using reverse substitution we can find our values for .

Otherwise, the depressed quartic can be solved by means of a method discovered by Lodovico Ferrari. Once the depressed quartic has been obtained, the next step is to add the valid identity

to equation (1), yielding

The effect has been to fold up the *u*^{4} term into a perfect square: (*u*^{2} + α)^{2}. The second term, α*u*^{2} did not disappear, but its sign has changed and it has been moved to the right side.

The next step is to insert a variable *y* into the perfect square on the left side of equation (2), and a corresponding 2*y* into the coefficient of *u*^{2} in the right side. To accomplish these insertions, the following valid formulas will be added to equation (2),

and

These two formulas, added together, produce

which added to equation (2) produces

This is equivalent to

The objective now is to choose a value for *y* such that the right side of equation (3) becomes a perfect square. This can be done by letting the discriminant of the quadratic function become zero. To explain this, first expand a perfect square so that it equals a quadratic function:

The quadratic function on the right side has three coefficients. It can be verified that squaring the second coefficient and then subtracting four times the product of the first and third coefficients yields zero:

Therefore to make the right side of equation (3) into a perfect square, the following equation must be solved:

Multiply the binomial with the polynomial,

Divide both sides by −4, and move the −*β*^{2}/4 to the right,

This is a cubic equation for *y*. Divide both sides by 2,

Equation (4) is a cubic equation nested within the quartic equation. It must be solved to solve the quartic. To solve the cubic, first transform it into a depressed cubic by means of the substitution

Equation (4) becomes

Expand the powers of the binomials,

Distribute, collect like powers of *v*, and cancel out the pair of *v*^{2} terms,

This is a depressed cubic equation.

Relabel its coefficients,

The depressed cubic now is

The solutions (any solution will do, so pick any of the three complex roots) of equation (5) are computed as (see Cubic equation)

where

and *V* is computed according to the two defining equations and , so

With the value for *y* given by equation (6), it is now known that the right side of equation (3) is a perfect square of the form

- (This is correct for both signs of square root, as long as the same sign is taken for both square roots. A ± is redundant, as it would be absorbed by another ± a few equations further down this page.)

so that it can be folded:

- .
- Note: If
*β*≠ 0 then*α*+ 2*y*≠ 0. If*β*= 0 then this would be a biquadratic equation, which we solved earlier.

- Note: If

Therefore equation (3) becomes

- .

Equation (7) has a pair of folded perfect squares, one on each side of the equation. The two perfect squares balance each other.

If two squares are equal, then the sides of the two squares are also equal, as shown by:

- .

Collecting like powers of u produces

- .
- Note: The subscript
*s*of and is to note that they are dependent.

- Note: The subscript

Equation (8) is a quadratic equation for *u*. Its solution is

Simplifying, one gets

This is the solution of the depressed quartic, therefore the solutions of the original quartic equation are

- Remember: The two come from the same place in equation (7''), and should both have the same sign, while the sign of is independent.

Given the quartic equation

its solution can be found by means of the following calculations:

If then

Otherwise, continue with

(either sign of the square root will do)

(there are 3 complex roots, any one of them will do)

As stated above, Cardano credited Ferrari as the first to discover one of these labyrinthine solutions. The equation he solved was:

which was already in depressed form. It has a pair of solutions that can be found with the set of formulas shown above.

If the coefficients of the quartic equation are real then the nested depressed cubic equation (5) also has real coefficients, thus it has at least one real root.

Furthermore the cubic function where P and Q are given by (5) has the properties that

- and

where α and β are given by (1).

This means that (5) has a real root greater than , and therefore that (4) has a real root greater than .

Using this root the term in (8) is always real, which ensures that the two quadratic equations (8) have real coefficients.^{[8]}

It could happen that one only obtained one solution through the seven formulae above, because not all four sign patterns are tried for four solutions, and the solution obtained is complex. It may also be the case that one is only looking for a real solution. Let *x*_{1} denote the complex solution. If all the original coefficients *A*, *B*, *C*, *D* and *E* are real — which should be the case when one desires only real solutions — then there is another complex solution *x*_{2}, which is the complex conjugate of *x*_{1}. If the other two roots are denoted as *x*_{3} and *x*_{4} then the quartic equation can be expressed as

but this quartic equation is equivalent to the product of two quadratic equations:

and

Since

then

Let

so that equation (9) becomes

Also let there be (unknown) variables *w* and *v* such that equation (10) becomes

Multiplying equations (11) and (12) produces

Comparing equation (13) to the original quartic equation, it can be seen that

and

Therefore

Equation (12) can be solved for *x* yielding

These two solutions are the desired real solutions if real solutions exist.

One can solve a quartic by factoring it into a product of two quadratics.^{[9]} Let

By equating coefficients, this results in the following set of simultaneous equations:

This can be simplified by starting again with a depressed quartic where , which can be obtained by substituting for , then , and:

It''s now easy to eliminate both and by doing the following:

If we set , then this equation turns into the resolvent cubic equation

which is solved elsewhere. Then:

The symmetries in this solution are easy to see. There are three roots of the cubic, corresponding to the three ways that a quartic can be factored into two quadratics, and choosing positive or negative values of for the square root of merely exchanges the two quadratics with one another.

The above solution shows that the quartic polynomial with a zero coefficient on the cubic term is factorable into quadratics with rational coefficients if and only if the resolvent cubic has a root which is the square of a rational; this can readily be checked using the rational root test.

The symmetric group S_{4} on four elements has the Klein four-group as a normal subgroup. This suggests using a **resolvent cubic** whose roots may be variously described as a discrete Fourier transform or aHadamard matrix transform of the roots; see Lagrange resolvents for the general method. Suppose r_{i} for i from 0 to 3 are roots of

If we now set

then since the transformation is an involution we may express the roots in terms of the four s_{i} in exactly the same way. Since we know the value s_{0} = -b/2, we really only need the values for s_{1}, s_{2} and s_{3}. These we may find by expanding the polynomial

which if we make the simplifying assumption that b=0, is equal to

This polynomial is of degree six, but only of degree three in z^{2}, and so the corresponding equation is solvable. By trial we can determine which three roots are the correct ones, and hence find the solutions of the quartic.

We can remove any requirement for trial by using a root of the same resolvent polynomial for factoring; if w is any root of (3), and if

then

We therefore can solve the quartic by solving for w and then solving for the roots of the two factors using the quadratic formula.

An alternative solution using algebraic geometry is given in (Faucette 1996), and proceeds as follows (more detailed discussion in reference). In brief, one interprets the roots as the intersection of two quadratic curves, then finds the three reducible quadratic curves (pairs of lines) that pass through these points (this corresponds to the resolvent cubic, the pairs of lines being the Lagrange resolvents), and then use these linear equations to solve the quadratic.

The four roots of the depressed quartic may also be expressed as the *x* coordinates of the intersections of the two quadratic equations i.e., using the substitution that two quadratics intersect in four points is an instance of Bézout''s theorem. Explicitly, the four points are for the four roots of the quartic.

These four points are not collinear because they lie on the irreducible quadratic and thus there is a 1-parameter family of quadratics (a pencil of curves) passing through these points. Writing the projectivization of the two quadratics as quadratic forms in three variables:

the pencil is given by the forms for any point in the projective line – in other words, where and are not both zero, and multiplying a quadratic form by a constant does not change its quadratic curve of zeros.

This pencil contains three reducible quadratics, each corresponding to a pair of lines, each passing through two of the four points, which can be done different ways. Denote these Given any two of these, their intersection is exactly the four points.

The reducible quadratics, in turn, may be determined by expressing the quadratic form as a 3×3 matrix: reducible quadratics correspond to this matrix being singular, which is equivalent to its determinant being zero, and the determinant is a homogeneous degree three polynomial in and and corresponds to the resolvent cubic.

or instead u could go to this website and chk it out urself...

http://en.wikipedia.org/wiki/Quartic_function#Solving_a_quartic_equation

8 years ago

if the equation is ax^2+bx+c=0 , then if b^2-4ac>o , roots are real & distinct......

if b^2-4ac =0 , then roots are real & equal..

if b^2-4ac<0 , then roots are complex & conjugate & distinct to each other....

8 years ago

are they both real? are they both imaginary?

Are they irrational?

take ax^{2}+bx+c=0

check (b^{2}-4ac)<>=0 depending on that you will get answers to above questions.

Rember if one root is imaginary that the other is its conjugate. If one root is irrational then the other is also.

8 years ago

you can use two things.either descartes rule of signs or discriminant

4 years ago

(-b^{2 }*(+or-) *under root of( b^{2}-4ac))/2a

where eqn = ax^{2} +bx+c=0

so substitute given coefficients and get the roots of equationss. then u will get two roots

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