• How to solve a biquadratic equation?

    Samiko
    Samiko
    September 18, 2014
    How to solve a biquadratic equation?

    Before you start solving a biquadratic equation, you should understand how it looks and how it differs from the classical quadratic equation. Equation of the ax4+ bx2+ c = 0 is called biquadratic with one variable (an algebraic equation of the fourth degree). To reduce the equation to a quadratic form and solve it through a discriminant, it is necessary to use the variable substitution:

    • ie: x2= t

    And then we have a standard equation of the form at2+ bt + c = 0

    The discriminant is calculated by the formula D = b2 - 4ac.

    1. In the case when D = 0, the equation has one single root t1= -b / 2a, and from here we get the desired solution of our equation x = sqrt (t1).
    2. If D> 0, the equation has two roots t1= (-b + sqrt (D)) / 2a and t2= (-b - sqrt (D)) / 2a. Do not forget about the variable entered, and we get the final solution x1,2= sqrt (t1) and x3,4= sqrt (t2)

    Important note: if any of the t valuesi<0, then for D = 0 the initial biquadratic solution has no real roots, and for D> 0, there is at most one single real root.

    Using the Viet theorem

    Good to know: in the case when we have a reduced quadratic equation (coefficient at t2= 1), the Viet theorem is applicable, and the search for a solution is minimized:

    • t1+ t2= -b
    • t1* t2= c

    Consider an example:

    • x4- 3x2+ 2 = 0

    using the substitution variable x2= t, we reduce the quadratic equation to the form t2- 3t; + 2 = 0.

    • D = (-3)2- 4*1*2 = 1.

    Roots of a quadratic equation t1= 2, t2= 1.

    Given the introduced change of variable, we obtain the solution of the desired biquadratic equation: t1= sqrt (2); t2= -sqrt (2); t3= 1; t4= -1.

    We can apply the Viet theorem to this task, since the coefficient of the variable with the highest power is 1:

    • t1+ t2= 3
    • t1* t2= 2

    From here t1= 2, t2= 1. As we can see, the roots of the quadratic equation in both cases coincide, which means that the solution of the biquadratic equation will be the same.

    In this article, we considered a special case of solving a biquadratic equation that is solved no more complicatedly than the classical quadratic equation.


    Related news


    How to find the cube diagonal
    Top 10 most popular fashion brands in the world
    How to choose a cooler
    What not to do on Good Friday
    What people need