How to Solve Biquadratic Equations?

Biquadratic equation, also known as “quartic equation without odd degree terms”, is used widely in algebra to simplify big polynomials in a systematic manner. The conventional “trial and error” technique to find the factors of a long polynomial equation is time-consuming. Hence, all the equations that satisfy the condition of not having odd degree terms can be solved using a biquadratic method. An equation in the form az4+bz2+c is called a biquadratic equation.  Solving a biquadratic equation is an easy mathematical approach.     

Method for solving biquadratic equation

After checking whether a polynomial equation can be converted to a biquadratic equation, i.e. no odd degree terms, the next step in the mathematical strategy is to transform the given polynomial equation into a biquadratic equation. The method of supposition is applied for the purpose. It is used for solving biquadratic equation to obtain the required roots. The obtained roots help study the nature of polynomials and the associated mathematical or real-world problems. In the supposition method, a variable is supposed. It is then substituted in the place of the actual variable in the polynomial. The dummy variable is selected so that it is with a specific degree to the actual variable. 

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Illustration 1

Consider the equation y=3t4+9t2-1. This is the biquadratic equation, two of whose roots are simplified below. Here, t is the actual polynomial. Now, suppose a variable such that the equation gets converted into biquadratic form. Suppose t2=x. Note that any such variable can be used. Further, substituting it in the equation, 

3(t2t2)+9t2-1=0

3(xx)+9x-1=0

3x2+9x-1=0

Solving the derived form to find the factors, the value of x is obtained. 

x = 0.1072 and x = -3.1072

After finding the answer in x terms, putting the actual value t2 back in the equation is the final step. 

t2 = 0.1072 and t2 = -3.1072

To find the value of t, take square roots on both sides, 

t = 0.3274. Thus, the biquadratic equation two, whose roots are t = 0.3274, is simplified. Usually, solving a biquadratic equation gives four roots, but there can be cases where it gives two roots only. 

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Illustration 2

Consider the equation t4-t2-2=0. Here, t is the actual polynomial. Now, suppose a variable such that the equation gets converted into biquadratic form. Suppose t2=z. Note that any such variable can be used. Further, substituting it in the equation, 

z2-z-2=0

Solving the derived form to find the factors, the value of z is obtained. 

z = 2 and z = -1

After finding the answer in z terms, putting the actual value t2 back in the equation is the final step. 

t2 = 2 and t2 = -1

To find the value of t, take square roots on both sides, 

t = 2

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Conclusion

The polynomial equation is of the form px4+qx3+rx2+sx+t=0. Here, x is the variable in the equation of degree 4. p, q, r, and s are the coefficient. t is the constant. In this equation, when the value of q and s is equal to zero. It takes the form px4+rx2+t=0. In case the value of q and s is not equal to zero, then the method of solving biquadratic equation cannot be used.