# Solving a Balanced Equation

In solving a B-square equation in IR, we must substitute its variable into a 2nd degree equation. Notice now the procedure that should be used.

Practical sequence:

• Replace x4 by y2 (or any other squared unknown) and x2 by y.

• Solve the equation ay2 + by + c = 0.

• Determine the square root of each of the roots (y'e y ") of the equation ay2 + by + c = 0.
These two relationships tell us that each positive root of the equation ay2 + by + c = 0 gives rise to two symmetrical roots for the square: the negative root gives no real root to it.

Examples:

• Determine the roots of the equation x4 - 13x2 + 36 = 0.
Solution:
Replacing x4 by y2 and x2 by y we have:
y2 - 13y + 36 = 0
By solving this equation we get:
y '= 4 and y "= 9
As x2= y, we have:

So we have for truth set: V = {-3, -2, 2, 3}.

• Determine the roots of the equation x4 + 4x2 - 60 = 0.
Solution:Replacing x4 by y2 and x2 by y we have:
y2 + 4y - 60 = 0
By solving this equation we get:
y '= 6 and y "= -10
As x2= y, we have:

So we have for the truth set:.

• Determine the sum of the roots of the equation .
Solution:We use the following device:

Like this:
y2 - 3y = -2
y2 - 3y + 2 = 0
y '= 1 and y "= 2
Substituting y, we determine:

Therefore, the sum of the roots is given by:

## Resolution of equations of the form: ax2n + bxno + c = 0

This kind of equation can be solved in the same way as the square one. For that, we replace xno by y getting:

ay2 + by + c = 0, which is an equation of the 2nd degree.

Example:

• Solve equation x6 + 117x3 - 1.000 = 0.
Solution:
Doing x3= y, we have:
y2 + 117y - 1,000 = 0

Solving the equation, we get:
y '= 8 and y "= - 125
So:

Thus, V = {-5,2}.

Next: Composition of the Balanced Equation