Binomial equations
Solving binomial equations

Solving equations reducible to quadratic
Binomial equations
An equation of the form  axn ± b = 0,  a > 0,  b > 0 and n is a natural number is called the binomial equation.
Solving binomial equations
 By using substitution the equation transforms to or y n  ± 1 = 0.
Recall that the expression on the left side can be factorized, for example:
y3  + 1 = (y + 1) · (y2 - y + 1)
y3  - 1 = (y - 1) · (y2 + y + 1)
y4  + 1 = (y2 + 1)2 - 2y2 = (y2  + 1 -Ö2 y) · (y2  + 1 + Ö2 y)
y4  - 1 = (y2 - 1) · (y2 + 1) = (y - 1) · (y + 1) · (y2 + 1)
y5  + 1 = (y + 1) ·  (y4 - y3 + y2 - y + 1)
y5  - 1 = (y - 1) ·  (y4 + y3 + y2 + y + 1)
y6  - 1 = ( y3  - 1) ·  (y3  + 1) = (y - 1) · (y + 1) · (y2 + y + 1) · (y2 - y + 1) and so on.
Example:  Solve the binomial equation  8x3 - 27  = 0.
 Solution:   Let substitute
To obtain the values of the original variable plug the solutions into the substitution  x = (3/2)y
Example:   Solve the binomial equation  x4 + 81 = 0.
 Solution:   By substituting obtained is
To obtain the values of the original variable plug the solutions into the substitution  x = 3y
Equations reducible to quadratic form - biquadratic equations
A bi-quadratic equation is said to be reducible to quadratic if the variable factor of the leading term is the square of the variable factor in the middle term.
Solving bi-quadratic equations or equations reducible to quadratic
Thus, a biquadratic equation
ax4 + bx2 + c = 0   we can write   a(x2)2 + bx2 + c = 0
and solve as the quadratic equation in the unknown x2 using the substitution x2 = y.
Example:   Solve the biquadratic equation  3x4 - 4x2 + 1 = 0.
Solution:   By substituting  x2 = y  we get the quadratic equation
3y2 - 4y + 1 = 0
To obtain the values of the original variable plug the solutions into the substitution  x2 = y
An equation of the form  ax2n + bxn + c = 0, where n is a natural number greater than 2, we can reduce to quadratic using the substitution xn = y.
So, obtained is the quadratic equation  ay2 + by + c = 0  the solutions of which are,  y1 and  y2
Then, to obtain the values of the original variable, we plug the solutions into the substitution xn = y.
Therefore,  xn = y1 and  xn = y2  are the two binomial equations
xn -  y1 = 0  and   xn -  y2 = 0
that have 2n solutions.
Example:   Solve the equation  x6 - 7x3 - 8 = 0  that is reducible to quadratic form.
Solution:   By substituting  x3 = y  we get the quadratic equation
y2 - 7y  - 8 = 0
To obtain the values of the original variable plug the solutions into the substitution  x3 = y
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