Symmetric equations of degree 4 have the form: ax4 + bx3 + x2 + bx + a = 0, a ≠ 0.
To solve a symmetric equation of degree 4, you need to proceed as follows.
Since x=0 is not a root of the equation, we divide both parts of the equation b x2. We get
ax2 + bx + + b/x + a/x2 = 0.
After grouping and factoring out common factors, the equation takes the form
a(x2 + 1/x2 )+ + b(x + 1/x )= 0.
Let's replace the variable t = x + 1/x, then t2 = x2 + 2 + 1/x2, therefor, t2 - 2 = x2 + 1/x2.
After substituting a new variable into the equation, we obtain a quadratic equation for t
a(t2 - 2) + bt + = 0 <=> at2 + bt + - 2a = 0
It remains to solve the resulting quadratic equation and reverse the substitution of variables.
In the same way, it is possible to solve equations of the form ax4 + bx3 + x2 - bx + a = 0, a ≠ 0.
Let's consider the solution of quartic symmetric equations using examples.
Examples of solving quartic symmetric equations
Example 1.
Solve the equation 3x4 + 5x3 - 2x2 + 5x + 3 = 0.
Solution.
After dividing both parts of the equation by x2, grouping the corresponding terms of the equation and factoring out the common factors, we obtain
3x2 + 5x - 2 + 5/x + 3/x2 = 0 <=> 3(x2 + 1/x2) + 5(x + 1/x) - 2 = 0
Let's replace the variable t = x + 1/x, then t2 = x2 + 2 + 1/x2, hence, t2 - 2 = x2 + 1/x2.
After substituting a new variable into the equation, we obtain a quadratic equation for t:
3(t2 - 2) + 5t - 2 = 0 <=> 3t2 + 5t - 8= 0.
We find its roots using well-known formulas
t1 = -8/3; t2 = 1.
It remains to make a reverse substitution and solve the resulting equations:
x + 1/x = -8/3 or x + 1/x = 1.
Multiplying both parts of these equations by x, we get two quadratic equations
3x2 + 8x + 3 = 0 or x2 - x + 1.
We find their roots:
x1,2 = (-4±√7)/3; x3,4 = (1±i√3)/2.
Answer: (-4-√7)/3, (-4+√7)/3, (1-i√3)/2, (1+i√3)/2.
Example 2.
Solve the equation 2x4 - 5x3 - x2 + 5x + 2 = 0.
Solution.
After dividing both parts of the equation by x2, grouping the corresponding terms of the equation and factoring out the common factors, we obtain
2x2 - 5x - 1 + 5/x + 2/x2 = 0 <=> 2(x2 + 1/x2) - 5(x - 1/x) - 1 = 0
Let's replace the variable t = x - 1/x, then t2 = x2 - 2 + 1/x2, hence, t2 + 2 = x2 + 1/x2.
After substituting a new variable into the equation, we obtain a quadratic equation for t:
2(t2 + 2) - 5t - 1 = 0 <=> 2t2 - 5t + 3 = 0.
We find its roots using well-known formulas
t1 = 3/2; t2 = 1.
It remains to make a reverse substitution and solve the resulting equations:
x - 1/x = 1 or x - 1/x = 3/2.
Multiplying both parts of these equations by x, we get two quadratic equations
x2 - x - 1 = 0 or 2x2 - 3x - 2 = 0.
We find their roots:
x1,2 = (1±√5)/2; x3,4 = (-3±5)/4.
Answer: -2, (1-√5)/2, 1/2, (1+√5)/2.
