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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