Solving Quartic Equations

Instructor: David Karsner
Quartic equations are polynomials that have a degree of four. There are several different means of solving quartic equations such as factoring, treating them like quadratics, use of a graphing calculator, and the rational root theorem.

Quartic Equations

Linear functions such as 2x-1=0 are easy to solve using inverse operations. Quadratic equations such as x2+5x+6 can be solved using the quadratic formula and breaking it down into linear factors. The polynomials of a higher order than two become more difficult to solve. Quartic equations are polynomials that have a degree of four, meaning the largest exponent is a four.

You would begin to solve quartic equations by setting it equal to zero. There will be four complex (real and imaginary) solutions, since it has a degree of four, to every quartic equation. Not every quartic equation will have four real roots. It could have 0, 1, 2, 3, or 4 real roots and imaginary roots making up the total of four. This lesson will show you several possible methods for solving quartic equations.

Difference of Squares

If the quartic equation is also a difference of squares; then it can be factored just like a difference of squares can be factored. A difference of squares comes in the form of a2-b2 and factors like (a+b)(a-b). For example x4-81=0 would factor like (x2+9)(x2-9)=0. The (x2-9) will factor again to (x-3)(x+3). We now have (x2+9)(x-3)(x+3)=0. This quartic has solutions at x=-3,and 3. The (x2+9) factor will never equal zero over real numbers so there will be two complex solutions.


When they Look Like Quadratics

Sometimes quartic equations can look like quadratic equations and have three terms. If a quartic has a term raised to the forth power, a term raised to the second power, and a constant; you can substitute x2 with another variable and then treat it like a quadratic.

Quadratics are polynomials that have a degree of two and can be solved with a variety of methods like factoring, completing the square, or using the quadratic formula.

Let's look at this example:


Notice it has three terms, a fourth power, a square, and a constant.

Replace x2 with the variable r.

Remeber that this means r2=x4

So we have 6r2-35r+50=0.

This quadratic will solve by factoring to give us (2r-5)(3r-10)=0.

Our solutions would be r=5/2 and 10/3.

Remember that r=x2.

That gives us x2=5/2 and x2=10/3.

One more step gives us x equal to the positive and negative square root of 5/2 and 10/3.

All four of the solutions in this quartic are real, irrational numbers.


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